diff --git a/chapter2.tex b/chapter2.tex new file mode 100644 index 0000000..f41e93c --- /dev/null +++ b/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}