Update output

chapter2.tex

@@ -0,0 +1,378 @@
+%! TeX root: ./main.tex
+
+
+\begin{qanda}
+  \question
+    Let $\cO_X$ be the pre-sheaf of smooth real-valued functions on $X = \bR^n$.
+    For any $p \in \bR^n$,
+    show that $\cO_p$, defined as a ring of germs of smooth functions,
+    is a local ring with maximal ideal $\fkm_p$,
+    the ideal of germs vanishing at $p$.
+  \answer
+    Let $\sigma \in \cO_p$ be some germ over $p$,
+    represented by a smooth function $\func{f}{U}{\bR}$ with $U \ni p$ open,
+    and $f$ not vanishing at $p$.
+    (Henceforth we will denote this by $\sigma = [f]_p$.)
+
+    Then since $f$ is smooth and non-vanishing at $p$,
+    there is some open nbd $V \subseteq U$ of $p$ supporting $f$.
+    Note that $[f|_V]_p = [f]_p$, by definition of germ equivalence.
+    Thus, we can define $g \in \cO_V$ by $g(x) = f^{-1}(x)$,
+    and this is well defined.
+
+    Finally, consider the germ $[g]_p \in \cO_p$.
+    Since $g \cdot f|_V = 1$, we have $[f]_p \cdot [g]_p = 1 \in \cO_p$.
+
+    Thus every germ over $p$ that does not vanish at $p$ is invertible,
+    and $\fkm_p$ is the unique maximal ideal as desired.
+
+  % Question 2.2.A
+  \question
+    Let $(X, \cTop_X)$ be a topological space,
+    with $\cTop_X$ considered as the poset category of open sets of $X$.
+    Show that a presheaf of sets over $X$ is precisely a contravariant functor
+    from $\cTop_X$ to $\Set$
+  \answer
+    The data of a functor $\func{\cF}{\cTop_X^{op}}{\Set}$
+    is a set $\cF(U)$ associated to each open set $U \in \cTop_X$,
+    and a function $\cF_{V,U}\colon \cF(V) \to \cF(U)$
+    for each inclusion $U \subseteq V$.
+    This is precisely the same data as a pre-sheaf $\cF$ of sets over $(X, \cTop_X)$.
+
+    For this data to form a functor is to require the identity $U \subseteq U$
+    to map to the identity function,
+    and to require the composition $U \subseteq V \subseteq W$
+    to map to $\cF_{W, U} = \cF{V, U} \circ \cF{W, V}$.
+    Again, these are precisely the same conditions required for $\cF$ to form a presheaf.
+
+  % Question 2.2.B
+  \question 
+    Show that the following are pre-sheaves on $\cC$, but not sheaves.
+    \begin{enumerate}
+      \item Bounded functions.
+      \item Holomorphic functions admitting a holomorphic square root.
+    \end{enumerate}
+  \answer
+    \begin{enumerate}
+      \item Define $\cF\colon \cTop_\bC^{op} \to \Set$ by sending each complex open set 
+        $U \subseteq \cC$ to the set of bounded functions on $U$.
+        The restriction maps $\cF_{V,U}\colon \cF(V) \to \cF(U)$ are true restrictions
+        $f \mapsto f|_U$.
+        Function restriction respects composition, i.e. $\big(f|_V\big)|_U = f|_U$,
+        hence $\cF$ defined in this way is a functor, and thus a pre-sheaf.
+
+        To see that it is not a sheaf,
+        note that an unbounded function may be bounded over every set in some open cover.
+        (Perhaps bounded functions never form a sheaf on a non-compact space?)
+
+        For example, the norm function over $\cC$ is bounded on restriction to any open cover
+        of $\bC$ by bounded opens, but is not bounded over all of $\cC$.
+      \item Now define $\cF\colon \cTop_\bC^{op} \to \Set$
+        by sending each complex open $U$ to
+        the set of holomorphic functions with holomorphic square root.
+        Again, send each inclusion $U \subseteq V$ to the true restriction map.
+        Since being holomorphic is a local property, it is preserved by restriction to
+        open subsets, and thus this is well-defined.
+        (Also note that the restriction of a square root will be a square root for the restriction.)
+
+        However, we may consider the identity map $z \mapsto z$.
+        This does not have a holomorphic square root (or even a continuous square root),
+        however, over any open set that does not contain a chosen ray from the origin,
+        the restriction of $z$ has a square root (a chosen branch cut).
+        Thus, we can cover $\cC$ with open sets where the restriction of $z$
+        has a homolorphic square root, but we cannot glue those sections together.
+
+        Thus, this does not form a sheaf over $\cC$.
+    \end{enumerate}
+  
+  % Question 2.2.C
+  \question
+    Interpret the identity and gluing axioms of a sheaf $\cF$
+    as saying that $\cF(\cup_{i \in I} U_i)$ is a certain limit.
+  \answer
+    Consider a sheaf of sets $\cF$ over some space $X$.
+    Let $U$ be open in $X$, with some open cover $\{U_i\}_{i \in I}$
+    for some indexing set $I$.
+
+    Then, consider the power set $\cP(I)$ as a poset category by inclusion,
+    with $J \to J'$ when $J \subseteq J'$.
+    Consider the diagram of sets $D_i\colon \cP(I) \to \Set$
+    sending each subset of indices $J$ to $\cF\big(\bigcap_{j \in J} U_j\big)$,
+    and each inclusion $J \to J'$ to the restriction map.
+    \[
+      \begin{tikzcd}
+        J \ar[d] & \cF\big(\bigcap_{j \in J} U_j\big) \ar[d] \\
+        J' & \cF\big(\bigcap_{j \in J'} U_j \big)
+      \end{tikzcd}
+    \]
+    The bottom of this diagram then looks like
+    \[
+      \begin{tikzcd}
+        \cF(U_i) \ar[rd] & & \cF(U_j) \ar[ld] \ar[rd] & & \cF(U_k) \ar[ld] \\
+                         & \cF(U_i \cap U_j) \ar[rd] & & \cF(U_j \cap U_k) \ar[ld] \\
+                         & & \cF(U_i \cap U_j \cap U_k) & &
+      \end{tikzcd}
+    \]
+    We claim that $\cF(U)$ is the limit in $\Set$ over this diagram.
+
+    The cone is given by the restriction maps $\cF(U) \to \cF(U_J)$ for each $J \subseteq I$.
+    By the definition of a pre-sheaf, these commute with the diagram functions,
+    which are all also restriction maps.
+
+    To see that this is the universal cone, consider another cone $\phi_i\colon Z \to \cF(U_i)$.
+    \[
+      \begin{tikzcd}
+        & & \color{orange} Z \ar[lldd, orange] \ar[dd, bend right, orange] \ar[rrdd, orange] \\
+        & & \cF(U) \ar[lld] \ar[d] \ar[rrd] & & \\
+        \cF(U_i) \ar[rd] & & \cF(U_j) \ar[ld] \ar[rd] & & \cF(U_k) \ar[ld] \\
+                         & \cF(U_i \cap U_j) & & \cF(U_j \cap U_k)
+      \end{tikzcd}
+    \]
+    For each $z \in Z$, denote $z_i = \phi_i(z)$.
+    Then since these sections agree on restriction, $z_i |_{U_i \cap U_j} = z_j |_{U_i \cap U_j}$
+    for each $i, j \in I$,
+    gluing induces some section $\phi(z) \in \cF(U)$ such that for all $i \in I$, $\phi(z)|_{U_i} = z_i$.
+
+    Define $\phi\colon Z \to \cF(U)$ by $z \mapsto \phi(z)$ as so.
+    Finally, an application of identity shows that this is unique,
+    since if $\varphi\colon Z \to \cF(U)$ is any other map commuting with the restriction maps,
+    then $\varphi(z)|_{U_i} = \phi(z)|_{U_i}$ for all $i \in I$, and hence $\varphi = \phi$.
+
+  % Exercise 2.2.D
+  \question
+    \begin{enumerate}
+      \item Verify that smooth functions, continuous functions, real-analytic functions,
+        and real-valued functions on a manifold or on $\bR^n$ form a shead.
+      \item Show that real-valued continuous functions on open sets of a topological space
+        form a sheaf.
+    \end{enumerate}
+  \answer
+    ...
+    Seems boring.
+
+    Yeah, the pre-sheaves are given by the states sets.
+    Restriction is given by true restriction.
+    All of the properties are local in the sense that they both
+    respect restriction, and may be determined by checking them on any cover.
+    Real functions always glue together uniquely from their restrictions
+    since they are determined pointwise.
+
+  % Exercise 2.2.E
+  \question
+    Let $S$ be any set, and
+    let $\cF(U)$ be the set of maps $U \to S$ which are locally constant.
+    Show this is a sheaf.
+
+    Equivalently, let $\cF(U)$ be the maps $U \to S$ which are continuous
+    when $S$ is endowed with the discrete topology
+    so that every subset is continuous.
+  \answer
+    We take the following definition of locally constant.
+    \[
+      \cF(U) = \{\func{f}{U}{S} \mid \forall p \in S, f^{-1}(p) \in \cTop_X \}
+    \]
+    Since this assigns to each open set a family of functions on it,
+    with the restriction maps being the standard restriction maps of functions,
+    the pre-sheaf structure and identity are both trivial.
+
+    To observe identity, consider some cover $U_i \hookrightarrow U$,
+    then take sections $f, g \colon U \to S$ such that $f_i = g_i$ for all $i \in I$
+    (where $f_i \ceq f|_{U_i}$ and $g_i \ceq g|_{U_i}$).
+    Then for any $x \in U$, $x \in U_i$ for some $i$, and $f(x) = f_i(x) = g_i(x) = g(x)$.
+
+    To show gluing, take some family $f_i \in \cF(U_i)$ such that
+    they agree on intersections, that is,
+    \[
+      \cndall i, j \in I, f_i|_{U_i \cap U_j} = f_j|_{U_i \cap U_j}.
+    \]
+    Since these are functions on $U_i$ which agree on intersections, they glue together
+    into a function on $U$.
+    \[
+      f\colon U \to S, \qquad x \mapsto f_i(x) \qqtext{where} x \in U_i.
+    \]
+    It remains to show that this is locally constant.
+    Take any $p \in S$.
+    Then 
+    \begin{align*}
+      f^{-1}(p) &= \{x \in U \mid f(x) = p\} \\
+                &= \{x \in U \mid \forall i \in I, p \in U_i \implies f_i(x) = p\} \\
+                &= \bigcup_{i \in I} f_i^{-1}(p)
+    \end{align*}
+    Which is a union of open sets since each $f_i$ is locally constant.
+    Thus, $f$ is locally constant, and the compatible $f_i$ sections glue together.
+
+    Thus, the locally constant functions form a sheaf.
+
+  % Exercise 2.2.F
+  \question
+    Let $X, Y$ be any topological spaces.
+    Show that continuous maps to $Y$ forms a sheaf of sets on $X$.
+  \answer
+    Again, the pre-sheaf structure is immediate
+    because we are assigning to each $U$, a family of functions on $U$,
+    with the standard restriction maps.
+
+    Identity is also immediate, by the same reason as the previous exercise.
+
+    Gluing is almost immediate since continuity is determined locally.
+    Let $U_i \hookrightarrow U$ be an open cover of some open $U \subseteq X$.
+    Then let $f_i \in \cF(U_i)$ be continuous functions to $Y$ agreeing on intersections
+    in the usual way.
+    Define $f\colon U \to Y$ as in the previous question,
+    with $f(x) = f_i(x)$ for $i \in I$ such that $x \in U_i$.
+    Well defined since the functions agree on intersections.
+
+    Then $f$ is continuous since for each $x \in U$, it has a neighbourhood $U_i$
+    where it is continuous.
+    Explicitly, if $V \subseteq Y$ is open, then
+    $f^{-1}(V) = \bigcup_{i \in I} f_i^{-1}(V)$
+    which is open by assumption.
+
+  % Exercise 2.2.G
+  \question
+    Let $X, Y$ be topological spaces.
+    \begin{enumerate}
+      \item Let $\mu\colon Y \to X$ be continuous.
+        Show that section of $\mu$ form a sheaf.
+        Explicitly, to each $U$ open in $X$,
+        assign the family of continuous maps $\func{s}{U}{Y}$
+        such that the following commutes.
+        \[
+          \begin{tikzcd}
+            U \ar[r, "s"] \ar[rr, bend right, "id|_U"'] & Y \ar[r, "\mu"] & X
+          \end{tikzcd}
+        \]
+      \item If $Y$ additionally has the structure of a topological group,
+        show that the continuous maps to $Y$ form a sheaf of groups.
+    \end{enumerate}
+  \answer
+  \begin{enumerate}
+    \item We first confirm that restriction is well defined.
+      Take some section $s\colon U \to Y$, and some open subset $U' \hookrightarrow U$.
+      Then $s|_{U'}$ is defined by precomposing $s$ with the inclusion.
+        \[
+          \begin{tikzcd}
+            U \ar[r, "s"] \ar[rr, bend left, "id|_U"] & Y \ar[r, "\mu"] & X \\
+            U' \ar[u] \ar[ru, "s|_{U'}"']
+          \end{tikzcd}
+        \]
+      Thus $\mu \circ s|_{U'} = (id|_U)|_{U'} = id|_{U'}$.
+      This shows that restriction of sections of $\mu$ is well defined,
+      and thus sections of $\mu$ do form a pre-sheaf.
+
+      Identity is as usual trivial since we are working with functions, determined pointwise.
+
+      For gluing, take a family $s_i\colon U_i \to Y$ which agree on intersections,
+      and then define $\func{s}{U}{Y}$ to be the unique function
+      agreeing with the $s_i$ on restriction. Then we have
+      \[
+        \begin{tikzcd}[row sep = small]
+          & U_i \cap U_j \ar[ld] \ar[rd] & \\
+          U_i \ar[dd, "s_i"'] \ar[rd] & & U_j \ar[ld] \ar[dd, "s_j"] \\
+                              & U \ar[ld, "s"] \ar[rd, "s"'] & \\
+          Y \ar[d, "\mu"'] & & Y \ar[d, "\mu"] \\
+          X \ar[rr, equal] & & X
+        \end{tikzcd}
+      \]
+      It remains to show that $s$ is a section.
+      Take any $x \in U$, then $x \in U_i$ for some $i \in I$,
+      and $\mu \circ s (x) = \mu \circ s_i (x) = x$.
+      Thus, yes, the compatible $s_i$ glue together into a section $s$,
+      and we have shown that sections of $\mu$ form a sheaf
+    \item Now assume that $Y$ is a topological group.
+      That is, that it has a continuous multiplication operation $\func{\times}{Y \times Y}{Y}$,
+      and the inverse $\func{(-)^{-1}}{Y}{Y}$ is also continuous.
+      We have already shown that continuous maps to $Y$ form a group.
+      It remains to show that the sets of sections form a group structure,
+      and that restriction is a group homomorphism.
+
+      To see the first, take two sections $f,g \colon U \to Y$.
+      Since these are continuous maps, there is an induced map $(f, g) \colon U \to Y \times Y$.
+      We may then post-compose this with multiplication to get $f\times g \colon U \to Y \times Y \to Y$.
+      \[
+        \begin{tikzcd}
+          & & Y \\
+          U \ar[rru, bend left, "f"] \ar[rrd, bend right,  "g"] \ar[r, dashed, "{(f, g)}"] & Y \times Y \ar[ru] \ar[rd] \ar[rr, "\times"] & & Y \\
+                                     & & Y
+        \end{tikzcd}
+      \]
+      Similarly, we may define an inverse function (under this operation) as
+      \[
+        \begin{tikzcd}
+          U \ar[r, "f"] \ar[rr, bend right, "\bar{f}"'] & Y \ar[r, "(-)^{-1}"] & Y\\
+        \end{tikzcd}
+      \]
+      Of course, these are just applying the operation and inverse point-wise,
+      but these constructions make it clear that $f \times g$ and $\bar{f}$ are well-defined
+      continuous maps.
+
+      To see that restriction is a group homomorphism, note that it is the same as precomposing the first diagram above with the relevant inclusion.
+      \[
+        \begin{tikzcd}
+          & & & Y \\
+          U' \ar[r] \ar[rrru, bend left, "f|_{U'}"] \ar[rrrd, bend right, "g|_{U'}"']
+          & U \ar[rru, bend left, "f"'] \ar[rrd, bend right,  "g"] \ar[r, dashed, "{(f, g)}"]
+          & Y \times Y \ar[ru] \ar[rd] \ar[rr, "\times"] & & Y \\
+          &                            & & Y
+        \end{tikzcd}
+      \]
+      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.
+
+\end{qanda}