654 lines
28 KiB
TeX
654 lines
28 KiB
TeX
%! TeX root: ./main.tex
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\begin{Lemma} \label{lem:subdiag-colim}
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Let $A\colon I \to \cC$ be a diagram in a category $\cC$,
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and let $B\colon J \to \cC$ be another diagram factoring through $A$.
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\[
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\begin{tikzcd}
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I \ar[r, "A"] & \cC \\
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J \ar[u, "\alpha"] \ar[ru, "B"']
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\end{tikzcd}
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\]
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Then the factoring $\alpha$ induces a map on the colimits, if they exist.
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\[
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\colim_{j \in J} B(j) \longrightarrow \colim_{i \in I} A(i)
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\]
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\end{Lemma}
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\begin{proof}
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Consider any co-cone $Z \in C$ over $A$ in $\cC$.
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We can write such a co-cone as a natural transformation
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$\eta$ from $A$ to the composition $I \to * \to \cC$,
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where $*$ is the discrete single object category,
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$* \to \cC$ selects the object $Z$,
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and $I \to *$ of course takes all objects to the single object.
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\[
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\begin{tikzcd}
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* \ar[rd, "Z"] \\
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I \ar[u] \ar[r, "A"'] & \cC
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\end{tikzcd}
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\]
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Note that the components of such a transformation are $\func{\eta_i}{A(i)}{Z}$,
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satisfying, for each $i \to j$ in $I$,
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\[
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\begin{tikzcd}
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A(i) \ar[r, "\eta_i"] \ar[d] & Z \ar[d, equals] \\
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A(j) \ar[r, "\eta_j"] & Z
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\end{tikzcd}
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\]
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This is precisely the definition of a co-cone.
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However, we can then take the composition $\nu = \eta * id_{\alpha}$,
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which naturally describes $Z$ as a co-cone over $B$.
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\[
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\begin{tikzcd}
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& * \ar[rd, "Z"] \\
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J \ar[r, "\alpha"'] \ar[ru] & I \ar[u] \ar[r, "A"'] & \cC
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\end{tikzcd}
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\]
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Thus, if $Z$ is any co-cone over $A$,
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and the colimit over $B$ exists,
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then there is a unique map from said colimit to $Z$ by universality.
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Taking $Z$ to be the co-cone over $A$ proves the lemma.
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\end{proof}
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\begin{Lemma} \label{lem:colim-composition}
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Let $A, B: I \to \cC$ be two diagrams in a category $\cC$,
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and let $\eta \colon A \implies B$ be a natural transformation between them.
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Then $\eta$ induces a map on the colimits, if they exist.
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\[
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\colim_{i \in I} B(i) \longrightarrow \colim_{i \in I} A(i)
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\]
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\end{Lemma}
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\begin{proof}
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Let $Z$ be any co-cone over the diagram $B$,
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written as a natural transformation as in the previous lemma.
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\[
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\begin{tikzcd}[row sep=large]
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& * \ar[rd, "Z"] \\
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I \ar[ru]
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\ar[rr, bend left, "A"{anchor=center, fill=white, name=A}]
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\ar[rr, bend right, "B"'{name=B}]
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\ar[from=B.north-|A, shorten=4pt, to=A, Rightarrow, "\eta"]
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\ar[from=A, ur, Rightarrow, shorten=2pt, "\varepsilon"]
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& & C\\
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\end{tikzcd}
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\]
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The composition $\varepsilon \circ \eta$ then describes $Z$
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as a co-cone over the diagram $B$.
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If we choose $Z$ to be the colimit over $A$,
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and assume the colimit of $B$ also exists,
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then we get the map desired by the lemma.
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\end{proof}
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\begin{qanda}
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\question
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Let $\cO_X$ be the pre-sheaf of smooth real-valued functions on $X = \bR^n$.
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For any $p \in \bR^n$,
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show that $\cO_p$, defined as a ring of germs of smooth functions,
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is a local ring with maximal ideal $\fkm_p$,
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the ideal of germs vanishing at $p$.
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\answer
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Let $\sigma \in \cO_p$ be some germ over $p$,
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represented by a smooth function $\func{f}{U}{\bR}$ with $U \ni p$ open,
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and $f$ not vanishing at $p$.
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(Henceforth we will denote this by $\sigma = [f]_p$.)
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Then since $f$ is smooth and non-vanishing at $p$,
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there is some open nbd $V \subseteq U$ of $p$ supporting $f$.
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Note that $[f|_V]_p = [f]_p$, by definition of germ equivalence.
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Thus, we can define $g \in \cO_V$ by $g(x) = f^{-1}(x)$,
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and this is well defined.
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Finally, consider the germ $[g]_p \in \cO_p$.
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Since $g \cdot f|_V = 1$, we have $[f]_p \cdot [g]_p = 1 \in \cO_p$.
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Thus every germ over $p$ that does not vanish at $p$ is invertible,
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and $\fkm_p$ is the unique maximal ideal as desired.
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% Question 2.2.A
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\question
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Let $(X, \cTop_X)$ be a topological space,
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with $\cTop_X$ considered as the poset category of open sets of $X$.
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Show that a presheaf of sets over $X$ is precisely a contravariant functor
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from $\cTop_X$ to $\Set$
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\answer
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The data of a functor $\func{\cF}{\cTop_X^{op}}{\Set}$
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is a set $\cF(U)$ associated to each open set $U \in \cTop_X$,
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and a function $\cF_{V,U}\colon \cF(V) \to \cF(U)$
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for each inclusion $U \subseteq V$.
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This is precisely the same data as a pre-sheaf $\cF$ of sets over $(X, \cTop_X)$.
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For this data to form a functor is to require the identity $U \subseteq U$
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to map to the identity function,
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and to require the composition $U \subseteq V \subseteq W$
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to map to $\cF_{W, U} = \cF{V, U} \circ \cF{W, V}$.
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Again, these are precisely the same conditions required for $\cF$ to form a presheaf.
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% Question 2.2.B
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\question
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Show that the following are pre-sheaves on $\cC$, but not sheaves.
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\begin{enumerate}
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\item Bounded functions.
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\item Holomorphic functions admitting a holomorphic square root.
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\end{enumerate}
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\answer
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\begin{enumerate}
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\item Define $\cF\colon \cTop_\bC^{op} \to \Set$ by sending each complex open set
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$U \subseteq \cC$ to the set of bounded functions on $U$.
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The restriction maps $\cF_{V,U}\colon \cF(V) \to \cF(U)$ are true restrictions
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$f \mapsto f|_U$.
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Function restriction respects composition, i.e. $\big(f|_V\big)|_U = f|_U$,
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hence $\cF$ defined in this way is a functor, and thus a pre-sheaf.
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To see that it is not a sheaf,
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note that an unbounded function may be bounded over every set in some open cover.
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(Perhaps bounded functions never form a sheaf on a non-compact space?)
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For example, the norm function over $\cC$ is bounded on restriction to any open cover
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of $\bC$ by bounded opens, but is not bounded over all of $\cC$.
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\item Now define $\cF\colon \cTop_\bC^{op} \to \Set$
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by sending each complex open $U$ to
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the set of holomorphic functions with holomorphic square root.
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Again, send each inclusion $U \subseteq V$ to the true restriction map.
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Since being holomorphic is a local property, it is preserved by restriction to
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open subsets, and thus this is well-defined.
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(Also note that the restriction of a square root will be a square root for the restriction.)
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However, we may consider the identity map $z \mapsto z$.
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This does not have a holomorphic square root (or even a continuous square root),
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however, over any open set that does not contain a chosen ray from the origin,
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the restriction of $z$ has a square root (a chosen branch cut).
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Thus, we can cover $\cC$ with open sets where the restriction of $z$
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has a homolorphic square root, but we cannot glue those sections together.
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Thus, this does not form a sheaf over $\cC$.
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\end{enumerate}
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% Question 2.2.C
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\question
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Interpret the identity and gluing axioms of a sheaf $\cF$
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as saying that $\cF(\cup_{i \in I} U_i)$ is a certain limit.
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\answer
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Consider a sheaf of sets $\cF$ over some space $X$.
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Let $U$ be open in $X$, with some open cover $\{U_i\}_{i \in I}$
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for some indexing set $I$.
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Then, consider the power set $\cP(I)$ as a poset category by inclusion,
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with $J \to J'$ when $J \subseteq J'$.
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Consider the diagram of sets $D_i\colon \cP(I) \to \Set$
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sending each subset of indices $J$ to $\cF\big(\bigcap_{j \in J} U_j\big)$,
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and each inclusion $J \to J'$ to the restriction map.
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\[
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\begin{tikzcd}
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J \ar[d] & \cF\big(\bigcap_{j \in J} U_j\big) \ar[d] \\
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J' & \cF\big(\bigcap_{j \in J'} U_j \big)
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\end{tikzcd}
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\]
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The bottom of this diagram then looks like
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\[
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\begin{tikzcd}
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\cF(U_i) \ar[rd] & & \cF(U_j) \ar[ld] \ar[rd] & & \cF(U_k) \ar[ld] \\
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& \cF(U_i \cap U_j) \ar[rd] & & \cF(U_j \cap U_k) \ar[ld] \\
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& & \cF(U_i \cap U_j \cap U_k) & &
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\end{tikzcd}
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\]
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We claim that $\cF(U)$ is the limit in $\Set$ over this diagram.
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The cone is given by the restriction maps $\cF(U) \to \cF(U_J)$ for each $J \subseteq I$.
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By the definition of a pre-sheaf, these commute with the diagram functions,
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which are all also restriction maps.
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To see that this is the universal cone, consider another cone $\phi_i\colon Z \to \cF(U_i)$.
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\[
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\begin{tikzcd}
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& & \color{orange} Z \ar[lldd, orange] \ar[dd, bend right, orange] \ar[rrdd, orange] \\
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& & \cF(U) \ar[lld] \ar[d] \ar[rrd] & & \\
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\cF(U_i) \ar[rd] & & \cF(U_j) \ar[ld] \ar[rd] & & \cF(U_k) \ar[ld] \\
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& \cF(U_i \cap U_j) & & \cF(U_j \cap U_k)
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\end{tikzcd}
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\]
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For each $z \in Z$, denote $z_i = \phi_i(z)$.
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Then since these sections agree on restriction, $z_i |_{U_i \cap U_j} = z_j |_{U_i \cap U_j}$
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for each $i, j \in I$,
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gluing induces some section $\phi(z) \in \cF(U)$ such that for all $i \in I$, $\phi(z)|_{U_i} = z_i$.
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Define $\phi\colon Z \to \cF(U)$ by $z \mapsto \phi(z)$ as so.
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Finally, an application of identity shows that this is unique,
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since if $\varphi\colon Z \to \cF(U)$ is any other map commuting with the restriction maps,
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then $\varphi(z)|_{U_i} = \phi(z)|_{U_i}$ for all $i \in I$, and hence $\varphi = \phi$.
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% Exercise 2.2.D
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\question
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\begin{enumerate}
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\item Verify that smooth functions, continuous functions, real-analytic functions,
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and real-valued functions on a manifold or on $\bR^n$ form a shead.
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\item Show that real-valued continuous functions on open sets of a topological space
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form a sheaf.
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\end{enumerate}
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\answer
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...
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Seems boring.
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Yeah, the pre-sheaves are given by the states sets.
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Restriction is given by true restriction.
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All of the properties are local in the sense that they both
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respect restriction, and may be determined by checking them on any cover.
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Real functions always glue together uniquely from their restrictions
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since they are determined pointwise.
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% Exercise 2.2.E
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\question
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Let $S$ be any set, and
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let $\cF(U)$ be the set of maps $U \to S$ which are locally constant.
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Show this is a sheaf.
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Equivalently, let $\cF(U)$ be the maps $U \to S$ which are continuous
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when $S$ is endowed with the discrete topology
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so that every subset is continuous.
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\answer
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We take the following definition of locally constant.
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\[
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\cF(U) = \{\func{f}{U}{S} \mid \forall p \in S, f^{-1}(p) \in \cTop_X \}
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\]
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Since this assigns to each open set a family of functions on it,
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with the restriction maps being the standard restriction maps of functions,
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the pre-sheaf structure and identity are both trivial.
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To observe identity, consider some cover $U_i \hookrightarrow U$,
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then take sections $f, g \colon U \to S$ such that $f_i = g_i$ for all $i \in I$
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(where $f_i \ceq f|_{U_i}$ and $g_i \ceq g|_{U_i}$).
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Then for any $x \in U$, $x \in U_i$ for some $i$, and $f(x) = f_i(x) = g_i(x) = g(x)$.
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To show gluing, take some family $f_i \in \cF(U_i)$ such that
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they agree on intersections, that is,
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\[
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\cndall i, j \in I, f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}.
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\]
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Since these are functions on $U_i$ which agree on intersections, they glue together
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into a function on $U$.
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\[
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f\colon U \to S, \qquad x \mapsto f_i(x) \qqtext{where} x \in U_i.
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\]
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It remains to show that this is locally constant.
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Take any $p \in S$.
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Then
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\begin{align*}
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f^{-1}(p) &= \{x \in U \mid f(x) = p\} \\
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&= \{x \in U \mid \forall i \in I, p \in U_i \implies f_i(x) = p\} \\
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&= \bigcup_{i \in I} f_i^{-1}(p)
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\end{align*}
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Which is a union of open sets since each $f_i$ is locally constant.
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Thus, $f$ is locally constant, and the compatible $f_i$ sections glue together.
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Thus, the locally constant functions form a sheaf.
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% Exercise 2.2.F
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\question
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Let $X, Y$ be any topological spaces.
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Show that continuous maps to $Y$ forms a sheaf of sets on $X$.
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\answer
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Again, the pre-sheaf structure is immediate
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because we are assigning to each $U$, a family of functions on $U$,
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with the standard restriction maps.
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Identity is also immediate, by the same reason as the previous exercise.
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Gluing is almost immediate since continuity is determined locally.
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Let $U_i \hookrightarrow U$ be an open cover of some open $U \subseteq X$.
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Then let $f_i \in \cF(U_i)$ be continuous functions to $Y$ agreeing on intersections
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in the usual way.
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Define $f\colon U \to Y$ as in the previous question,
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with $f(x) = f_i(x)$ for $i \in I$ such that $x \in U_i$.
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Well defined since the functions agree on intersections.
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Then $f$ is continuous since for each $x \in U$, it has a neighbourhood $U_i$
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where it is continuous.
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Explicitly, if $V \subseteq Y$ is open, then
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$f^{-1}(V) = \bigcup_{i \in I} f_i^{-1}(V)$
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which is open by assumption.
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% Exercise 2.2.G
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\question
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Let $X, Y$ be topological spaces.
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\begin{enumerate}
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\item Let $\mu\colon Y \to X$ be continuous.
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Show that section of $\mu$ form a sheaf.
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Explicitly, to each $U$ open in $X$,
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assign the family of continuous maps $\func{s}{U}{Y}$
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such that the following commutes.
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\[
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\begin{tikzcd}
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U \ar[r, "s"] \ar[rr, bend right, "id|_U"'] & Y \ar[r, "\mu"] & X
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\end{tikzcd}
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\]
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\item If $Y$ additionally has the structure of a topological group,
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show that the continuous maps to $Y$ form a sheaf of groups.
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\end{enumerate}
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\answer
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\begin{enumerate}
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\item We first confirm that restriction is well defined.
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Take some section $s\colon U \to Y$, and some open subset $U' \hookrightarrow U$.
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Then $s|_{U'}$ is defined by precomposing $s$ with the inclusion.
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\[
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\begin{tikzcd}
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U \ar[r, "s"] \ar[rr, bend left, "id|_U"] & Y \ar[r, "\mu"] & X \\
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U' \ar[u] \ar[ru, "s|_{U'}"']
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\end{tikzcd}
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\]
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Thus $\mu \circ s|_{U'} = (id|_U)|_{U'} = id|_{U'}$.
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This shows that restriction of sections of $\mu$ is well defined,
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and thus sections of $\mu$ do form a pre-sheaf.
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Identity is as usual trivial since we are working with functions, determined pointwise.
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For gluing, take a family $s_i\colon U_i \to Y$ which agree on intersections,
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and then define $\func{s}{U}{Y}$ to be the unique function
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agreeing with the $s_i$ on restriction. Then we have
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\[
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\begin{tikzcd}[row sep = small]
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& U_i \cap U_j \ar[ld] \ar[rd] & \\
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U_i \ar[dd, "s_i"'] \ar[rd] & & U_j \ar[ld] \ar[dd, "s_j"] \\
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& U \ar[ld, "s"] \ar[rd, "s"'] & \\
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Y \ar[d, "\mu"'] & & Y \ar[d, "\mu"] \\
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X \ar[rr, equal] & & X
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\end{tikzcd}
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\]
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It remains to show that $s$ is a section.
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Take any $x \in U$, then $x \in U_i$ for some $i \in I$,
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and $\mu \circ s (x) = \mu \circ s_i (x) = x$.
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Thus, yes, the compatible $s_i$ glue together into a section $s$,
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and we have shown that sections of $\mu$ form a sheaf
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\item Now assume that $Y$ is a topological group.
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That is, that it has a continuous multiplication operation $\func{\times}{Y \times Y}{Y}$,
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and the inverse $\func{(-)^{-1}}{Y}{Y}$ is also continuous.
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We have already shown that continuous maps to $Y$ form a group.
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It remains to show that the sets of sections form a group structure,
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and that restriction is a group homomorphism.
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To see the first, take two sections $f,g \colon U \to Y$.
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Since these are continuous maps, there is an induced map $(f, g) \colon U \to Y \times Y$.
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We may then post-compose this with multiplication to get $f\times g \colon U \to Y \times Y \to Y$.
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\[
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\begin{tikzcd}
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& & Y \\
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U \ar[rru, bend left, "f"] \ar[rrd, bend right, "g"] \ar[r, dashed, "{(f, g)}"] & Y \times Y \ar[ru] \ar[rd] \ar[rr, "\times"] & & Y \\
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& & Y
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\end{tikzcd}
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\]
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Similarly, we may define an inverse function (under this operation) as
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\[
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\begin{tikzcd}
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U \ar[r, "f"] \ar[rr, bend right, "\bar{f}"'] & Y \ar[r, "(-)^{-1}"] & Y\\
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\end{tikzcd}
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\]
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Of course, these are just applying the operation and inverse point-wise,
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but these constructions make it clear that $f \times g$ and $\bar{f}$ are well-defined
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continuous maps.
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To see that restriction is a group homomorphism, note that it is the same as precomposing the first diagram above with the relevant inclusion.
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\[
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\begin{tikzcd}
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& & & Y \\
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U' \ar[r] \ar[rrru, bend left, "f|_{U'}"] \ar[rrrd, bend right, "g|_{U'}"']
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& U \ar[rru, bend left, "f"'] \ar[rrd, bend right, "g"] \ar[r, dashed, "{(f, g)}"]
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& Y \times Y \ar[ru] \ar[rd] \ar[rr, "\times"] & & Y \\
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& & & Y
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\end{tikzcd}
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\]
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|
By universal property of the product, we have $(f, g)|_{U'} = (f|_{U'}, g|_{U'})$,
|
|
and thus $f|_{U'} \times g|_{U'} = (f \times g)|_{U'}$.
|
|
|
|
Thus $\Hom(-, Y)$ forms a sheaf of groups on $X$ as desired.
|
|
\end{enumerate}
|
|
|
|
% Question 2.2.H
|
|
\question
|
|
Let $\func{\pi}{X}{Y}$ be a continuous map.
|
|
Assume $\cF$ is a pre-sheaf on $X$.
|
|
Then define a new pre-sheaf $\pi_*\cF$ on $Y$,
|
|
called the pushforward of $\cF$ by $\pi$,
|
|
by $\pi_*\cF(V) = \cF(\pi^{-1}(V))$, for open $V$ in $Y$.
|
|
|
|
Show that $\pi_* \cF$ is a pre-sheaf,
|
|
and show it is a sheaf when $\cF$ is.
|
|
\answer
|
|
Note that since $\pi$ is a continuous map,
|
|
$\pi^{-1}$ defines an inclusion preserving map from open sets of $Y$ to open sets of $X$.
|
|
That is, $\pi^{-1}$ is a functor $\cTop_Y \to \cTop_X$.
|
|
$\pi_*\cF$ is precisely the composition of $\cF$ with (the dual of) this functor.
|
|
\[
|
|
\begin{tikzcd}
|
|
\cTop_Y^{op} \ar[r, "\pi^{-1}_{op}"] \ar[rr, bend right, "\pi_*\cF"'] & \cTop_X^{op} \ar[r, "\cF"] & \Set
|
|
\end{tikzcd}
|
|
\]
|
|
Thus, by the contravariant functor description, $\pi_*\cF$ is a pre-sheaf.
|
|
|
|
Now, assume that $\cF$ is furthermore a sheaf.
|
|
We shall show that $\pi_*\cF$ is also a sheaf.
|
|
Take some open $V$ in $Y$, and some cover $V_i \hookrightarrow V$ as usual.
|
|
|
|
To show identity, take $f, g \in \pi_*\cF(V) = \cF(\pi^{-1}(V))$,
|
|
such that
|
|
\[
|
|
(\pi_*\cF)_{V, V_i}(f) = \cF_{\pi^{-1}(V), \pi^{-1}(V_i)}(f)
|
|
= \cF_{\pi^{-1}(V), \pi^{-1}(V_i)}(g)
|
|
= (\pi_*\cF)_{V, V_i}(g) .
|
|
\]
|
|
Then by identity of $\cF$, we have $f = g$ as sections over $\pi^{-1}(V)$,
|
|
and then hence as sections over $V$.
|
|
|
|
To show gluing, take $f_i \in \cF(\pi^{-1}(V_i))$,
|
|
agreeing when restricted to $V_i \cap V_j$. That is,
|
|
\[
|
|
\cF_{\pi^{-1}(V_i), \pi^{-1}(V_i \cap V_j)}(f_i)
|
|
= (\pi_*\cF)_{V_i, V_i \cap V_j}(f_i)
|
|
= (\pi_*\cF)_{V_i, V_i \cap V_j}(g_i)
|
|
= \cF_{\pi^{-1}(V_i), \pi^{-1}(V_i \cap V_j)}(f_i).
|
|
\]
|
|
Noting that since $\{V_i\}$ is a cover of $V$,
|
|
then $\{\pi^{-1}(V_i)\}$ is a cover of $\pi^{-1}(V)$,
|
|
and $\pi^{-1}(V_i \cap V_j) = \pi^{-1}(V_i) \cap \pi^{-1}(V_j)$,
|
|
then gluing in $\cF$ yields a section $f \in \cF(\pi^{-1}(V)) = \pi_*\cF(V)$.
|
|
|
|
This, if $\cF$ is a sheaf on $X$,
|
|
then the pushforward of $\cF$ over any continuous map $X \to Y$
|
|
is also a sheaf on $Y$, as desired.
|
|
|
|
% Question 2.2.I
|
|
\question
|
|
Let $\func{\pi}{X}{Y}$ be a continuous map of spaces,
|
|
and $\cF$ be a sheaf on $X$.
|
|
if $\pi(p) = q$ for some $p \in X$,
|
|
describe the natural morphism $(\pi_*\cF)_q \to \cF_p$.
|
|
\answer
|
|
We first define the map explicitly.
|
|
Take some germ $[f]_q$ in $(\pi_*\cF)_q$,
|
|
with $f \in (\pi_*\cF)(V)$ a section over some open $V \ni q$.
|
|
Then by the definition of the sections in the pushforward sheaf,
|
|
$f \in (\pi_*\cF)(V) = \cF(\pi^{-1}(V))$,
|
|
thus $f$ is also an element of the set of sections of $\cF$ over the open nbd
|
|
$\pi^{-1}(V)$ of $p$. Thus, we can define
|
|
$[f]_p \in \cF_p$, the germ of $f$ over $p$.
|
|
The induced map on stalks may then be expressed simply as $[f]_q \mapsto [f]_p$.
|
|
|
|
Alternatively, we can induce the map using the universal property of the stalks.
|
|
\[
|
|
\begin{tikzcd}
|
|
\cTop_{Y}^{op}|_{q} \ar[r, "\pi^{-1}_q"] \ar[d] & \cTop_{X}^{op}|_{p} \ar[d] \\
|
|
\cTop_{Y}^{op} \ar[r, "\pi^{-1}"] & \cTop_{X}^{op} \ar[r, "\cF"] & \Set
|
|
\end{tikzcd}
|
|
\]
|
|
Here $\cTop_Y^{op}|_q$ denotes the full subcategory of $\cTop_Y^{op}$
|
|
consisting of open sets containing $q$, and analagously for $\cTop_X^{op}|_p$.
|
|
|
|
The stalk $\cF_p$ is then the colimit over the diagram
|
|
$\cTop^{op}_X|_p \to \cTop^{op}_X \to \Set$,
|
|
and the stalk $(\pi_*\cF)_q$ is the colimit over the diagram
|
|
$\cTop_Y^{op}|_q \to \cTop_Y^{op} \to \cTop_X^{op} \to \Set$.
|
|
|
|
Lemma \ref{lem:subdiag-colim} then induces the desired map by universal property of colimit.
|
|
|
|
% Question 2.2.J
|
|
\question
|
|
If $(X, \cO_X)$ is a ringed space,
|
|
and $\cF$ is an $\cO_X$-module, describe how for each $p \in X$,
|
|
$\cF_p$ is an $\cO_{X, p}$-module.
|
|
\answer
|
|
We shall describe the action explicitly.
|
|
Let $p \in X$, and consider $\cF_p$.
|
|
We already know this retains the structure of an abelian group,
|
|
since the category of abelian groups has colimits.
|
|
To define the action of $\cO_{X, p}$,
|
|
let $[f]_p \in \cO_{X, p}$ and $[a]_p \in \cF_p$
|
|
where $f$ and $a$ are sections of their respective sheaves
|
|
over some shared neighbourhood $U \ni p$ (restricting if necessary).
|
|
|
|
Then define $[f]_p \cdot [a]_p$ to be $[f\cdot a]_p$
|
|
using the action of $\cO_X(U)$ on $\cF(U)$.
|
|
To see that this is well defined, let $[f]_p = [f']_p$ and $[a]_p = [a']_p$
|
|
where $f', a'$ are defined over another neighbourhood $V$ of $p$.
|
|
|
|
Since they determine the same germ, there are some $W_1, W_2$ around $p$
|
|
where $f|_{W_1} = f'|_{W_2}$ and $a|_{W_1} = a'_{W_2}$.
|
|
Defining $W = W_1 \cap W_2$, we can finally put everything into the same neighbourhood,
|
|
so that $f|_W = f'|_W$ and $a|_W = a'|_W$.
|
|
|
|
Then the compatibility condition for the $\cO_X$-module $\cF$ says that
|
|
$(f \cdot a)|_W = f|_W \cdot a|_W = f'_W \cdot a'|_W = (f' \cdot a')|_W$.
|
|
Thus $[f \cdot a]_p = [f'\cdot a']_p$, and this definition of action on stalks
|
|
is well defined.
|
|
|
|
Finally, note that all the axioms of module action may be determined with
|
|
a finite number of elements, and we may thus check them
|
|
by taking representatives on some shared neighbourhood of $p$,
|
|
where the axioms hold by the definition of an $\cO_X$-module.
|
|
|
|
% Question 2.3.A
|
|
\question
|
|
Let $\phi\colon \cF \to \cG$ be a morphism of pre-sheaves on $X$,
|
|
and let $p \in X$.
|
|
Then describe the induced morphism of stalks $\phi_p\colon \cF_p \to \cG_p$.
|
|
|
|
This defines the \emph{stalkification functor at $p$}
|
|
$\Sets_X \to \Sets$, where $\Sets_X$ is the category of
|
|
sheaves of sets over $X$.
|
|
\answer
|
|
This follows immediately from Lemma \ref{lem:colim-composition},
|
|
since $\phi$ induces a natural transformation between the stalk diagrams
|
|
for $\cF_p$ and $\cG_p$.
|
|
|
|
% Question 2.3.B
|
|
\question
|
|
Let $\pi\colon X \to Y$ be a continuous map of topological spaces.
|
|
Show that the pushforward induces a functor
|
|
$\pi_*\colon \Sets_X \to \Sets_Y$.
|
|
\answer
|
|
Recall that $\pi_*\cF$ is pre-composition with $\pi^{-1}_{op}$.
|
|
Then we can describe the desired functor as
|
|
\[
|
|
\begin{tikzcd}
|
|
\cF \ar[d, Rightarrow, "\eta"] \ar[r, mapsto] &
|
|
\pi_*\cF = \cF \circ \pi^{-1}_{op} \ar[d, Rightarrow, "\eta \circ id_{\pi^{-1}_{op}}"] \\
|
|
\cG \ar[r, mapsto] & \pi_*\cG = \cG \circ \pi^{-1}_{op}
|
|
\end{tikzcd}
|
|
\]
|
|
That is, on objects it acts as pre-composition by $\pi^{-1}_{op}$,
|
|
and on morphisms it acts as pre-composition by (the natural transformation) $id_{\pi^{-1}_{op}}$
|
|
|
|
% Question 2.3.C
|
|
\question
|
|
Let $\cF$ and $\cG$ be two sheaves over $X$.
|
|
Define a pre-sheaf of sets $\cHom(\cF, \cG)$
|
|
by lettings the set of sections over each open $U$ be
|
|
\[
|
|
\cHom(\cF, \cG)(U) \ceq \Mor(\cF|_U, \cG|_U).
|
|
\]
|
|
Show that this is a sheaf of set on $X$.
|
|
\answer
|
|
We first observe that for any $U$ open in $X$,
|
|
the sections of $\cHom(\cF, \cG)$ over $U$ are defined to be the set
|
|
of natural transformations $\eta$ in the diagram below.
|
|
The restriction of a section $\eta$ to some open $V \subseteq U$
|
|
is then a composition of $\eta$ with the (identity transformation on the) inclusion
|
|
of $\cTop_V^{op} \to \cTop_U^{op}$
|
|
\[
|
|
\begin{tikzcd}
|
|
& & \cTop_X^{op} \ar[rd, "\cF"] & \\
|
|
\cTop_V^{op} \ar[r] & \cTop_U^{op} \ar[ru] \ar[rd] & & \Set \\
|
|
& & \cTop_X^{op} \ar[ru, "\cG"'] \ar[uu, Rightarrow, shorten=5pt, "\eta"] &
|
|
\end{tikzcd}
|
|
\]
|
|
Pre-composition is functorial,
|
|
and so $\cHom(\cF, \cG)$ does form a presheaf of sets.
|
|
|
|
Next, we need to show that it is also a sheaf.
|
|
We first consider identity, in the usual way.
|
|
Let $U$ be open in $X$, and take an open cover $U = \bigcup_{i \in I} U_i$.
|
|
Let $\eta, \varepsilon\colon \cF|_U \Rightarrow \cG_U$,
|
|
such that $\eta|_{U_i} = \varepsilon|_{U_i}$ for all $i \in I$.
|
|
We claim that $\eta = \varepsilon$.
|
|
To see this, take any open $V \subseteq U$,
|
|
and define the cover $V_i = V \cap U_i$ of $V$. Then:
|
|
\[
|
|
\eta_{V_i} = (\eta|_{U_i})_{V_i} = (\varepsilon|_{U_i})_{V_i} = \varepsilon_{V_i}.
|
|
\]
|
|
We have the following diagram.
|
|
\[
|
|
\begin{tikzcd}
|
|
\cF|_U(V) \ar[r, "\eta_V"] \ar[d] & \cG|_U(V) \ar[d] \\
|
|
\cF|_U(V_i) \ar[r, "\eta_{V_i}", "\varepsilon_{V_i}"'] & \cG|_U(V_i) \\
|
|
\cF|_U(V) \ar[u] \ar[r, "\varepsilon_{V}"] & \cG|_U(V) \ar[u]
|
|
\end{tikzcd}
|
|
\]
|
|
Here the vertical maps are restriction maps, and both squares commute.
|
|
Chasing any section $f \in \cF|_U(V)$, we see that
|
|
\[
|
|
(\eta_V(f))|_{V_i} = \eta_{V_i}(f|_{V_i}) = \varepsilon_{V_i}(f|_{V_i}) = (\varepsilon_V(f))|_{V_i}.
|
|
\]
|
|
Since $V_i$ forms a cover of $V$, identity in the sheaf $\cG$ then tells us $\eta_V = \varepsilon_V$.
|
|
Since $V$ was arbitrary, this means that $\eta = \varepsilon$ as desired.
|
|
|
|
Finally, we demonstrate gluing.
|
|
Let $U_i$ form a cover of $U$ as above,
|
|
and let $\eta^i\colon \cF|_{U_i} \Rightarrow \cG|_{U_i}$ in $\cHom(\cF, \cG)(U_i)$,
|
|
such that they agree when restricted to intersections, that is,
|
|
\[
|
|
\forall W \subseteq U_i \cap U_j, \qquad
|
|
\eta^i_W = \big(\eta^i|_{U_i \cap U_j}\big)_W = \big(\eta^j|_{U_i \cap U_j}\big)_W
|
|
= \eta^j_W.
|
|
\]
|
|
Take any $V$ open in $U$, and as before define the cover $V_i = V \cap U_i$ of $V$.
|
|
We define a gluing $\eta\colon \cF|_U \Rightarrow \cG|_U$
|
|
by defining it on the arbitrary component $V$.
|
|
\[
|
|
\begin{tikzcd}
|
|
\cF(V) \ar[r, dashed, "\eta_V"] \ar[d] & \cG(V) \ar[d] \\
|
|
\cF(V_i) \ar[r, "\eta^i_{V_i}"] \ar[d] & \cG(V_i) \ar[d] \\
|
|
\cF(V_i \cap V_j) \ar[r, "\eta^i_{V_i \cap V_j}", "\eta^j_{V_i \cap V_j}"'] & \cG(V_i \cap V_j)
|
|
\end{tikzcd}
|
|
\]
|
|
Taking any section $f \in \cF(V) = \cF|_{U}(V)$,
|
|
define $g_i = \eta^i_{V_i}(f|_{V_i}) \in \cG(V_i)$.
|
|
Then
|
|
\[
|
|
g_i|_{V_i \cap V_j} =
|
|
\eta^i_{V_i}(f|_{V_i}) |_{V_i \cap V_j}
|
|
= \eta^i_{V_i \cap V_j}(f|_{V_i \cap V_j})
|
|
= \eta^j_{V_i \cap V_j}(f|_{V_i \cap V_j})
|
|
= \eta^j_{V_j}(f|_{V_j}) | _{V_i \cap V_j}
|
|
= g_j |_{V_i \cap V_j}.
|
|
\]
|
|
Thus $\{g_i\}_{i\in I}$ is a family of sections of $\cG$ over $V_i$
|
|
which agree upon restriction to intersections.
|
|
Since $\cG$ was assumed a sheaf, it satisfies gluing,
|
|
and there is thus some $g \in \cG(V)$ such that $g|_{V_i} = g_i$.
|
|
|
|
We define $\eta_V(f) = g$.
|
|
|
|
Thus, $\cHom(\cF, \cG)$ is a presheaf satisfying both identity and gluing, and is thus a sheaf,
|
|
as desired.
|
|
|
|
|
|
|
|
|
|
\end{qanda}
|