--- title: "Algebraic Geometry Problem Sets" subtitle: "Problem Set 2" author: - name: D. Zack Garza affiliation: University of Georgia email: dzackgarza@gmail.com date: Spring 2022 order: 2 --- # Problem Set 2 ## Problem 1 :::{.proposition title="1.1"} The global sections functor is left-exact. ::: :::{.proof} We'll use the fact that a sequence of sheaves is exact if and only if the induced sequence on stalks is exact. Given this, let $\mcf, \mcg, \mch$ be sheaves of abelian groups on $X$, and consider the diagram induced by restriction morphisms: \begin{tikzcd} 0 && \mcf && \mcg && \mch && 0 \\ \\ && {\mcf(U)} && {\mcg(U)} && {\mch(U)} \\ \\ 0 && {\mcf_p} && {\mcg_p} && {\mch_p} && 0 \arrow[from=1-1, to=1-3] \arrow["f", from=1-3, to=1-5] \arrow["g", from=1-5, to=1-7] \arrow[from=1-7, to=1-9] \arrow[from=5-1, to=5-3] \arrow["{f_p}", from=5-3, to=5-5] \arrow["{g_p}", from=5-5, to=5-7] \arrow[from=5-7, to=5-9] \arrow[from=1-7, to=3-7] \arrow[from=3-7, to=5-7] \arrow[from=3-5, to=5-5] \arrow[from=1-3, to=3-3] \arrow[from=3-3, to=5-3] \arrow["{f_U}", from=3-3, to=3-5] \arrow[from=1-5, to=3-5] \arrow["{g_U}", from=3-5, to=3-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTMsWzAsMCwiMCJdLFsyLDAsIlxcbWNmIl0sWzQsMCwiXFxtY2ciXSxbNiwwLCJcXG1jaCJdLFs4LDAsIjAiXSxbMiwyLCJcXG1jZihVKSJdLFsyLDQsIlxcbWNmX3AiXSxbNCwyLCJcXG1jZyhVKSJdLFs0LDQsIlxcbWNnX3AiXSxbNiwyLCJcXG1jaChVKSJdLFs2LDQsIlxcbWNoX3AiXSxbOCw0LCIwIl0sWzAsNCwiMCJdLFswLDFdLFsxLDIsImYiXSxbMiwzLCJnIl0sWzMsNF0sWzEyLDZdLFs2LDgsImZfcCJdLFs4LDEwLCJnX3AiXSxbMTAsMTFdLFszLDldLFs5LDEwXSxbNyw4XSxbMSw1XSxbNSw2XSxbNSw3LCJmX1UiXSxbMiw3XSxbNyw5LCJnX1UiXV0=) Note that we can take $U = X$ in this diagram. If the top sequence of sheaves is exact, there are isomorphisms of sheaves: - $\ker f = 0$ and - $\im f = \ker g$. :::{.claim} \[ \ker f_X = 0 ,\] making $f_X$ injective and yielding exactness at the first position. ::: :::{.proof title="?"} Since the presheaf $\ker f$ is in fact a sheaf, writing $\mathbf 0$ for the sheaf $U\mapsto 0$, we have \[ \ker\qty{ \mcf(X) \mapsvia{f_X} \mcg(X) } = (\ker f)(X) = \mathbf{0}(X) = 0 \] ::: :::{.claim} \[ \im f_X = \ker g_X ,\] yielding exactness at the middle position. ::: :::{.proof title="?"} $\im f_X \subseteq \ker g_X$ follows from a diagram chase: \begin{tikzcd} && \textcolor{rgb,255:red,92;green,92;blue,214}{a} && \textcolor{rgb,255:red,92;green,92;blue,214}{b } && \textcolor{rgb,255:red,92;green,92;blue,214}{\therefore g_X(b) = 0} \\ && {\mcf(X)} && {\mcg(X)} && {\mch(X)} \\ \\ 0 && {\mcf_p} && {\mcg_p} && {\mch_p} && 0 \\ && \textcolor{rgb,255:red,92;green,92;blue,214}{a_p} &&&& \textcolor{rgb,255:red,92;green,92;blue,214}{\tilde a = (g_p\circ f_p)(a_p) = 0 \, \forall p} \arrow[from=4-1, to=4-3] \arrow["{f_p}", from=4-3, to=4-5] \arrow["{g_p}", from=4-5, to=4-7] \arrow[from=4-7, to=4-9] \arrow[from=2-7, to=4-7] \arrow[from=2-5, to=4-5] \arrow[from=2-3, to=4-3] \arrow["{f_X}", from=2-3, to=2-5] \arrow["{g_X}", from=2-5, to=2-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=18pt}, dashed, maps to, from=1-3, to=5-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, dashed, maps to, from=5-3, to=5-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, dotted, maps to, from=1-3, to=1-5] \arrow[color={rgb,255:red,92;green,92;blue,214}, dotted, maps to, from=1-5, to=1-7] \arrow[color={rgb,255:red,92;green,92;blue,214}, curve={height=-24pt}, dotted, maps to, from=1-7, to=5-7] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) - Fix $b\in \im f_X \subseteq \mcg(X)$, then by surjectivity choose a lift $a\in \mcf(X)$. - Map $a$ along $\mcf(X)\to \mcf_p \to \mcg_p \to \mch_p$; by exactness the result is zero in $C_p$. - By commutativity of the diagram, mapping $a$ along $\mcf(X)\to \mcg(X)\to \mch(X) \to \mch_p$ also yields zero in $C_p$. - Write $\tilde b \da g_X(b)$; since the above argument holds for all $p\in X$, $g_X(b)$ is a section of $\mch$ that is zero in every stalk. Thus by the sheaf property for $\mch$, the section $g_X(b)$ must be zero, and $b\in \ker g_X$. Similarly, $\ker g_X \subseteq \im f_X$ follows from a diagram chase: - Fix $b\in \ker g_X$, so the image of $g_X$ in $\mch(X)$ is zero. Then its image $c_p$ along $\mcg(X) \to \mch(X) \to \mch_p$ is also zero. - By commutativity of the right square, $g_p(b_p) = 0$ and so $b_p\in \ker g_p = \im f_p$ by exactness of the bottom row. - Choose a lift $a_p\in \mcf_p$ along $f_p$, so $f_p(a_p) = b_p$ Since $a_p$ is a germ of $\mcf$, pick any global section $a\in \mcf(X)$ restricting to $a_p$ and making the square commute. - Since $f_X(a)_p = f_p(a_p) = b_p$ for all $p$, by uniqueness of gluing for $\mcg$ we have $f_X(a) = b$ and $b\in \im f_X$. ::: ::: :::{.proposition title="1.2"} Taking global sections may fail to be right-exact. ::: :::{.proof title="?"} Consider the following poset and its corresponding category of open sets: \begin{tikzcd} X &&&& {\Open(X)} \\ a &&&& {\ts{a,b,c}} \\ \\ b &&& {\ts{a,b}} && {\ts{b,c}} \\ \\ c &&& {\ts{b}} && {\ts{c}} \\ \\ &&&& \emptyset \arrow[from=8-5, to=6-6] \arrow[from=8-5, to=6-4] \arrow[from=6-4, to=4-6] \arrow[from=6-6, to=4-6] \arrow[from=6-4, to=4-4] \arrow[""{name=0, anchor=center, inner sep=0}, from=4-4, to=2-5] \arrow[from=4-6, to=2-5] \arrow["\leq", from=6-1, to=4-1] \arrow[""{name=1, anchor=center, inner sep=0}, "\leq", from=4-1, to=2-1] \arrow[shorten <=24pt, shorten >=24pt, Rightarrow, from=1, to=0] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzQsMSwiXFx0c3thLGIsY30iXSxbMywzLCJcXHRze2EsYn0iXSxbNSwzLCJcXHRze2IsY30iXSxbMyw1LCJcXHRze2J9Il0sWzUsNSwiXFx0c3tjfSJdLFs0LDcsIlxcZW1wdHlzZXQiXSxbMCwxLCJhIl0sWzAsMywiYiJdLFswLDUsImMiXSxbMCwwLCJYIl0sWzQsMCwiXFxPcGVuKFgpIl0sWzUsNF0sWzUsM10sWzMsMl0sWzQsMl0sWzMsMV0sWzEsMF0sWzIsMF0sWzgsNywiXFxsZXEiXSxbNyw2LCJcXGxlcSJdLFsxOSwxNiwiIiwwLHsic2hvcnRlbiI6eyJzb3VyY2UiOjIwLCJ0YXJnZXQiOjIwfX1dXQ==) Define the following two sheaves $\mcf, \mcg$ and a morphism between them: ![](figures/2022-02-18_10-18-05.png) > [Link to diagram](https://q.uiver.app/?q=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) Note that there are only three stalks to consider, none of which coincide with global sections, so we can take the sheaf morphism to be the identity on these to get a surjection on stalks. We then choose a non-surjective map $\mcf(X) \to \mcg(X)$ given by $(a, b) \mapsto (a, a+b)$, where e.g. the image does not contain the element $(1, 1)$. One can check that the individual diagrams for $\mcf$ and $\mcg$ commute, yielding a presheaf, and that existence and uniqueness of gluing hold for both. Moreover, all of the squares formed by the map $\mcf\to \mcg$ commute, so this does in fact yield a morphism of sheaves. ::: ## Problem 2 :::{.proposition title="?"} If a map $f:X\to Y$ between posets is continuous, it is order-preserving, i.e. if $x_1\leq x_2$ then $f(x_1) \leq f(x_2)$. ::: :::{.proof title="?"} Continuity can be checked on a basis, so let $U_b = \ts{y\in Y \st y\geq b}$ be a basic open upper set. Then $f$ is continuous iff $f\inv(U_a)$ is an open set in $X$. Being open means that for every $x_0 \in f\inv(U_a)$, $x_1\geq x_0\implies x_1\in f\inv(U_a)$. \[ f \text{ is continuous } &\iff \forall U \text{ open in } Y, \, f\inv(U) \text{ is open in } X \\ &\iff \forall U_a \text{ a basic open in } Y, \, f\inv(U_a) \text{ is open in } X \\ &\iff \forall a\in Y,\, \forall x_0 \in f\inv(U_a),\, x_1\geq x_0\implies x_1\in f\inv(U_a) \\ &\iff \forall a\in Y,\, \forall x_0 \in f\inv(U_a),\, x_1\geq x_0\implies f(x_1) \in U_a \\ &\iff \forall a\in Y,\, \forall x_0 \in f\inv(U_a),\, x_1\geq x_0\implies f(x_1) \geq a\\ &\iff \forall a\in Y,\, \forall x_0\in X \text{ s.t. } f(x_0) \geq a,\, x_1\geq x_0\implies f(x_1) \geq a .\] Now taking $x_0 = f\inv(a)$ for $a\in \im f$ yields \[ \implies \forall a\in \im f,\quad x_1 \geq f\inv(a) \implies f(x_1) \geq a .\] Relabeling $x_1 = f\inv(b)$, \[ &\implies \forall a\in \im f, \quad f\inv(b) \geq f\inv(a) \implies b \geq a \\ &\implies \forall \tilde a\in f\inv(Y), \quad \tilde b \geq \tilde a \implies f(\tilde b) \geq f(\tilde a) .\] ::: :::{.proposition title="?"} For $\mcf\in \Sh_X, \mcg \in \Sh_Y, \mch \in \Sh_U$ with $U \subseteq X$, $X \mapsvia{f} Y$, and $U \injectsvia{j} X$, - $f_* \mcf$ is no additional data - $f\inv \mcg = \ts{\mcg_{f(x_0)}, \phi_{f(x_0), f(x_1) } \st x_0, x_1\in X, x_0 \leq x_1}$. - $j_! \mch = ?$ ::: :::{.proof} We'll use that $\mcf\in \Sh(X, \Ab\Grp)$ is the same as the data of $\ts{\mcf_x, \phi_{xy}}$ where $\mcf_x$ is a collection of groups and $\phi_{xy}: \mcf_x\to \mcf_y$ are group morphisms for every $x\leq y$. Thus the values of a sheaf on posets are entirely determined by a functorial assignment of groups to the stalk at each point, i.e. an assignment of a group to each point. So it suffices to determine what the stalks of these three sheaves are. - For $f_*\mcf$, noting that \[ (f_* \mcf)(U_{\geq a}) = \mcf(f\inv(U_{\geq a})) = \cocolim_{x\in f\inv(U_{\geq a})} \mcf_x ,\] we see that this sheaf is completely determined by the data for $\mcf$. - For $f\inv\mcg$, we can use the fact that for any sheaf, there is a formula on stalks: \[ (f\inv \mcg)_p \cong \mcg_{f(p)} ,\] and so $f\inv \mcg$ is the data $\ts{\mcg_x, \psi_{xy}}$ for every $x\leq y$ with $x,y\in \im f$. - For $j_! \mch, \cdots ?$ > Attempts to approach this: the general definition involves sheafification, which seems hard to describe in general. On the other hand, I haven't been able to work out what the sheaf space for a poset should look like. ::: ## Problem 3 :::{.proposition title="?"} Let $\mcf\in \Sh(X)$ and let $\Et(\mcf) \mapsvia{\pi} X$ be its corresponding sheaf space, so $\mcf = \Sec_{\cts}(\pi)$, and let $\mcg = \Sec(\pi)$. Then \[ \mcg = \prod_{x\in X} x_* \mcf_x \] where $x: \ts{x} \injects X$ is the inclusion of a point and $\mcf \in \Sh(\ts{x})$ is regarded as a sheaf on a one-point space. ::: :::{.proof} We'll use the fact that as a set, $\Et(\mcf) = \coprod_{x\in X} \mcf_x$ is the coproduct of all of the stalks of $\mcf$. We can compute the sections of this sheaf as follows: \[ \mcg(U) &= \qty{ \prod_{x\in X} x_* \mcf_x}(U) \\ &= \prod_{x\in X} (x_* \mcf_x)(U) \\ &= \prod_{x\in X} \mcf_x(x\inv(U)) \\ &= \prod_{x\in X} \mcf_x\qty{ \begin{cases} \ts{x} & x\in U \\ \emptyset & x\not\in U. \end{cases} } \\ &= \prod_{x\in X} \begin{cases} \mcf_x & x\in U \\ 0 & x\not\in U. \end{cases} \\ &= \prod_{x\in U} \mcf_x .\] We can now simply regard $\mcg(U)$ as the set of set-valued functions $s: U\to \coprod_{x\in U} \mcf_x \subseteq \Et(\mcf)$ by setting $s(x) = \pi_x\qty{\prod_{x\in U} \mcf_x}$ to be the $x\dash$coordinate in the direct product, where $\pi_x: \prod_{x\in U}\mcf_x \to \mcf_x$ is projection onto the $x\dash$coordinate. On the other hand, the data of a set-valued section $s \in \Sec( \Et(\mcf) \mapsvia{\pi} U)$ is the following: for every $x\in X$, a choice of an element \[ s(x) \in \pi\inv(x) = \mcf_x \subseteq \Et(\mcf) ,\] with no other compatibility conditions, which is precisely the same as the set-valued functions specified by $\mcg(U)$ above. ::: :::{.proposition title="?"} The stalks $\mcg_p$ are given by \[ \mcg_p = \colim_{U\ni p} \prod_{x\in U} \mcf_x ,\] the direct limit of the product of stalks of $\mcf$ along neighborhoods of $p$. ::: :::{.proof} \[ \mcg_p &\da \colim_{U\ni p} \mcg(U) \\ &\da \colim_{U\ni p} \qty{ \prod_{x\in X} (\iota_x)_* \mcf_x }(U) \\ &= \colim_{U\ni p} \prod_{x\in X} \qty{ (\iota_x)_* \mcf_x} (U) \\ &\da \colim_{U\ni p} \prod_{x\in X} \mcf_x(\iota_x\inv(U)) \\ &= \colim_{U\ni p} \prod_{x\in X} \mcf_x\qty{ \begin{cases} \ts{x} & x\in U \\ \emptyset & \text{else}. \end{cases} } \\ &= \colim_{U\ni p} \prod_{x\in X} \qty{ \begin{cases} \mcf_x & x\in U \\ 0 & \text{else}. \end{cases} } \\ &= \colim_{U\ni p} \prod_{x\in U} \mcf_x .\] ::: :::{.proposition title="?"} There is an injective morphism of sheaves $\mcf \injects \mcg$. ::: :::{.proof} For every open $U \subseteq X$, define a map of sets on the function spaces: \[ \Psi_U: \Sec_{\cts}(\Et(\mcf) \mapsvia{\pi} U) &\to \Sec(\Et(\mcf) \mapsvia{\pi} U) \\ f &\mapsto f ,\] which does nothing more than a forgetful map that regards a continuous section as a set-valued section. This is evidently an injective map of sets, since if $f_1, f_2$ are continuous sections and $f_1 = f_2$ as set-valued functions, they continue to be equal when regarded as continuous sections, so $\Psi_U(f_1) = \Psi_U(f_2) \implies f_1 = f_2$. These $\Psi_U$ assemble to a morphism of sheaves $\Psi: \mcf\to \mcg$, and since $(\ker \Psi)^- = \mathbf{0}$ vanishes as a presheaf and the kernel presheaf is a sheaf, we have $\ker \Psi = \mathbf{0}$. :::