# Confusion about elements of the stalk of the direct image sheaf

$\DeclareMathOperator{\F}{\mathcal{F}}$Let $\F$ be a sheaf on $X$ and $\pi : X \rightarrow Y$ a continuous map. Then the the direct image sheaf $\pi_* \F$ is a sheaf on $Y$. An explicit definition of the stalk the sheaf $\F$ at point $p \in X$ is as follows: $$\F_p = \{ (f, U) \mid p \in U, \, \, f \in \F(U) \} / \sim$$ where $(f, U) \sim (g, V)$ if and only if there exists an open $W \subset U \cap V$ such that $f|_W = g|_W$. Now, the stalk of $\pi_* \F$ at point $q = \pi(p) \in Y$ is as follows: \begin{align*} (\pi_* \F)_q &= \{ (h, Z) \mid q \in Z, \, \, h \in \pi_* \F(Z) = \F(\pi^{-1}(Z)) \} / \sim \\ &\stackrel{\color{red}{?}}{=} \{ (h, \pi^{-1}(Z)) \mid p \in \pi^{-1}(Z), \, \, h \in \F(\pi^{-1}(Z)) \} / \sim \end{align*} but then since $h \in \F(\pi^{-1}(Z))$, it's e.g. a continuous function in some subset of $X$ (say the sheaves are of continuous functions). Hasn't something gone wrong here? Surely if I'm talking about the stalk of $\pi_* \F$, then shouldn't the function be in $Y$, not $X$? I know that there's a natural map between $(\pi_* \F)_q$ and $\F_p$ so this likely has something do with my question, but all I've done above is write out what the stalk $(\pi_* \F)_q$ is explicitly, and there already seems to be some contradiction regarding the function $h$.

Thank you for any help.

Edit: Having looked at the above, I don't think it's valid me jumping to the second line because this is no longer correct as an explicit realisation of the stalk of $\pi_* \F$, since inside the curly brackets I'm taking sections in $\F$ whereas I should be taking them in $\pi_* \F$.

However I'm still confused; doesn't $f \in \pi_*\F(U)$ mean that $f$ is, say, a continuous function in $Y$ (because $\pi_*\F$ is a sheaf over $Y$), but then simultaneously $f \in \F(\pi^{-1}(U))$ is a continuous function in $X$ because $\F$ is a sheaf over $X$? I feel I'm misunderstanding something really obvious here.

Edit 2: Maybe this isn't relevant. $\F$ being a sheaf on $X$ means its arguments should be open sets in $X$, but it doesn't actually say anything about what the sections have to be?

• Don't forget that for $p\in Y$ there may be several $p\in X$ with $\pi(x)=y$, (or perhaps none). Jul 25 '18 at 17:49
• Do you by any chance mean for $q \in Y$, there may be several $p \in X$ with $\pi(p)=q$? Also, how could there be none if we've said initially that $\pi(p)=q$? Jul 25 '18 at 17:53
• The direct image sheaf will still have a stalk at $q$ even if $\pi^{-1}(q)=\emptyset$, Jul 25 '18 at 17:54
• The stalks of $\pi_*\mathcal F$ are equivalence classes of certain sections of $\mathcal F$ over certain subsets of $X$ under a certain equivalence relation. All the "certain"s in that sentence could be replaced by "rather complicated". Anyway, these "rather complicated" objects are certainly too involved for me to readily visualise them. Jul 25 '18 at 18:11
• Your second edit addresses the material point here, IMO: I think your main misconception is that sections of a sheaf on $\mathcal{F}$ on a space $X$ need to have anything to do with $X$. (This is not an unreasonable confusion: many natural sheaves on a space $X$ will actually be related to $X$ in some way. But this need not be the case.) Jul 25 '18 at 21:07

Let $f:X\to Y$ be a continuous map of topological spaces, let $\mathcal{S}$ be the sheaf of continuous functions on $X$ with values in $\mathbb{K}$ with the natural topology, where $\mathbb{K}$ is either $\mathbb{R}$ the field of real numbers or $\mathbb{C}$ the field of complex numbers.
For exact: $$\forall U\subseteq X\,\text{open,}\,\mathcal{S}(U)=\{c:U\to\mathbb{K}\mid c\,\text{is continuous}\};$$ because $\mathcal{S}$ is a sheaf of rings on $X$, the pair $(X,\mathcal{S})$ is called ringed space.
By definition: $$\forall V\subseteq Y\,\text{open,}\,f_{*}\mathcal{S}(V)\stackrel{def.}{=}\mathcal{S}(f^{-1}(V)),$$ that is: $\pi_{*}\mathcal{S}$ is a sheaf of rings on $Y$; the elements of $f_{*}\mathcal{S}(V)$ can be identified with functions on $V$, with values in $\mathbb{K}$ which admit a factorization via $f$. In other words, $f_{*}\mathcal{S}$ is a subsheaf of the sheaf of functions on $Y$ with values in $\mathbb{K}$.
Remark. For clarity: $$\forall V\subseteq Y\,\text{open,}\,f_{*}\mathcal{S}(V)=\{d:V\to\mathbb{K}\mid\exists c\in\mathcal{S}(f^{-1}(V))\,\text{s.t.}\,d=c\circ f\},$$ and one can not state that $d$ is continuous on $V$. $\Diamond$
Let $$\widetilde{X}=X_{\displaystyle/(x\sim y\iff f(x)=f(y))},$$ that is $\widetilde{X}$ is the quotient set of $X$ where the points with the same image via $f$ are identied; let $\pi:X\to\widetilde{X}$ be the canonical projection, and let $\varphi:\widetilde{X}\to Y$ be the unique function such that $$f=\varphi\circ\pi.$$ One knows that $\varphi$ is an injective map; considering on $\widetilde{X}$ the quotient topology, one has that $\pi$ and $\varphi$ are continuous maps and $$f_{*}\mathcal{S}=(\varphi\circ\pi)_{*}\mathcal{S}=\varphi_{*}\left(\pi_{*}\mathcal{S}\right).$$ Because $f(X)=\varphi\left(\widetilde{X}\right)=Z$ is a subset of $Y$ and $\varphi$ is injective, one knows that: $$\forall y\in Y,\,\left(f_{*}\mathcal{S}\right)_y=\begin{cases} 0\iff y\notin\overline{Z}\\ \left(\pi_{*}\mathcal{S}\right)_z\iff\varphi(z)=y\in Z\\ ?\iff y\in\overline{Z}\setminus Z \end{cases}.$$ Example. Let $$X=\{a,b,c\},\,\mathcal{T}=\{\emptyset,\{b\},\{c\},\{b,c\},X\};$$ $\{a\}$ is a closed point and $\{b\}$ is an open point of $X$. Let $$\mathcal{F}(\emptyset)=\{0\},\,\mathcal{F}(\{b\})=\mathbb{Z},$$ this is a sheaf on $\{b\}$ with the topology of subspace of $(X,\mathcal{T})$; easily one has: $$i_{*}\mathcal{F}(X)=\mathbb{Z}\Rightarrow\left(i_{*}\mathcal{F}\right)_a=\mathbb{Z},\\ \left(i_{*}\mathcal{F}\right)_b=\mathbb{Z},\\ i_{*}\mathcal{F}(\{c\})=\{0\},i_{*}\mathcal{F}(\{b,c\})=\mathbb{Z}\Rightarrow\left(i_{*}\mathcal{F}\right)_c=\{0\}$$ where $i:\{b\}\hookrightarrow X$ is the inclusion and $a\in\overline{\{b\}}\setminus\{b\}$. Instead, let $\mathcal{G}$ be the analogous sheaf on $\{a\}$; one has that $j_{*}\mathcal{G}$ is the skyscraper sheaf on $X$ with stalk $\mathbb{Z}$ at $a$, where $j:\{a\}\to X$ is the inclusion. $\triangle$
So, also if one computes what is $\left(\pi_{*}\mathcal{S}\right)_z$ for any $z\in\widetilde{X}$, in general one can state nothing about $\left(f_{*}\mathcal{S}\right)_y$ for $y\in\overline{Z}\setminus Z$ a priori.