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I am working through the proof of proposition 5 in Section 5: Extension and restriction of a sheaf in FAC by Serre. FAC can be found in english here, and in particular this question arises on page 11, but I will also include the entire proposition and proof below, and I will also warn that Serre uses the etale space definition of a sheaf (much older), not the modern functor definition. The categories are equivalent. If you have only seen the modern definition this will seem strange. If anyone needs me to include certain definitions if they are unfamiliar with notation Serre uses, just let me know.

I feel this should be and easy proof to follow, but I am having trouble in some spots.

Definition: Let $X$ be a topological space, $Y$ its closed subspace and $\mathscr{F}$ a sheaf on $X$. We say $\mathscr{F}$ is concentrated on $Y$ if we have $\mathscr{F}_x = 0$ for all $x \in X-Y$.

Question 1 [answered!]: Why does he say "Y its closed subspace" as though it is unique? Was this just a weird bit with translating from french? I am just interpreting it as $Y is a closed subspace".

Proposition 5. If a sheaf $\mathscr{F}$ is concentrated on $Y$, the homomorphism $\rho_{Y}^{X}: \Gamma(X,\mathscr{F}) \to \Gamma(Y,\mathscr{F}(Y))$ is bijective (thus an isomorphism).

Remarks: $\mathscr{F}(Y)$ is the sheaf induced on $Y$ by $\mathscr{F}$. $\mathscr{F}_x$ is the stalk of $\mathscr{F}$ over the point $x$. $\Gamma(U,\mathscr{F})$ is the group of sections over $U$ with respect to the sheaf $\mathscr{F}$.

Proof: [copied verbatim] If a section of $\mathscr{F}$ over $X$ is zero over $Y$, it is zero everywhere else since $\mathscr{F}_{x} = 0$ if $x \notin Y$, which shows that $\rho_{Y}^{X}$ is injective. Conversely, let $s$ be a section of $\mathscr{F}(Y)$ over $Y$, and extend $s$ onto $X$ by putting $s(x) = 0$ for $x \notin Y$; the mapping $x \mapsto s(x)$ is clearly continuous on $X - Y$. On the other hand, if $x \in Y$ there exists a section $s'$ of $\mathscr{F}$ over an open neighborhood $U$ of $x$ for which $s'(x) = s(x)$. Since $s$ is continuous on $Y$ by assumption, there exists an open neighborhood $V$ of $x$, contained in $U$ and such that $s'(y) = s(y)$ for all $y \in V \cap Y$. But since $\mathscr{F}_y = 0$ for $y \notin Y$, we also have that $s'(y) = s(y)$ for $y \in V-(V \cap Y)$; hence $s$ and $s'$ coincide on $V$, which proves that $s$ is continuous in a neighborhood of $Y$, so it is continuous everywhere.

My interpretation / what I do understand: The injectivity part is fine, and on a big picture I understand what we're trying to do in surjectivity - we nominate the only possible candidate for a preimage (the extension of a section over $Y$ by 0). This candidate will easily satisfy the composition requirement with the local homeomorphism to be a section, so we need continuity. This is the interesting part of the result, because extensions of this sort are not always continuous for general topological settings, so this is a sheaf property. So we have this closed subspace $Y$ and some continuous map $s: Y \to \mathscr{F}(Y)$ and we form this new piece wise map (which I call) $s^{*}: X \to \mathscr{F}$ given by $$ s^{*}: x \mapsto \cases{ s(x), &x $\in$ Y\\ 0, &x $\notin$ Y }$$ I understand how Serre says that $s^{*}$ is continuous on $X-Y$, which is because the trivial global section is always continuous, and this is just a restriction of that. I also understand why it is justified when he says that the section $s'$ exists for some open subset $U$ with the property $s(x) = s'(x)$ for the $x \in Y$. It is after this that things are semi-familiar, but not entirely clear to me.

Uncertainties:

1) First of all why does he start his surjectivity proof by saying "conversely"? I have never thought of surjectivity as the converse of injectivity?

2) Picking up from where I left off with what I do understand, in Serres next line he asserts the existence of the open set $V$ contained in $U$ with the property that $s$ and $s'$ agree on $V \cap Y$. It seems like he is invoking the fact that any two sections with the property that $s(x) = s'(x)$ must agree on some open neighborhood, but the $V$ he invokes seems different than this? Once the existence of this $V$ is justified I understand why it follows that $s$ and $s'$ agree on $V$, but then I am again confused how to explain why it follows $s$ is continuous everywhere. It makes intuitive sense, because our problem spots are really just the boundary of $Y$? But for every single point of $Y$, we have an open set $V_y$ so that $s$ is continuous on $V_y$, then $s$ is also continuous on $Y^c$ so we can paste together all the $V_y$'s and $Y^c$ to get one big continuous function?

3) Does the above result actually rely on $Y$ being closed? Does it not hold for $Y$ open? Aside from the result, are there reasons that we define the concentrated sheaf definition only for closed $Y$? I know things can get weird with the direct limits when single points are closed, are we avoiding stuff like that?

Follow up:

After some time thinking about this, the only part that is really an issue now is how Serre gets the set $V$ with the property that it is contained in $U$ and $s$,$s'$ agree on $V \cap Y$.

Edit: I think I am close?

We know that for any point $a \in \mathscr{F}$ there is an open neighborhood $W$ of $a$ such that $\pi|_{W} W \to \pi(W)$ is a homeomorphism with $\pi(W)$ open in $X$. Take $ V = (s')^{-1}(W)$, then $V \subseteq U$ and $V \subseteq \pi(W)$.

Since $(\pi|_{W})^{-1}$ is a bijection then for each $w \in \pi(W)$ there is exactly one element, say $(f,w)$ of $\mathscr{F}_W$ in $W$. So if $v \in V = (s')^{-1}(W)$ then $s'(v) \in W$ and say $s'(v)=(g,v)$ but $(\pi|_{W})^{-1}(v) = (g,v)$ so $(\pi|_{W})^{-1}$ and $s'$ agree on $V$.

But then when I intersect with $Y$, why would $s$ and $s'$ agree?

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    $\begingroup$ I checked the french version and about your "question 1" it is indeed "Y is a closed subset of X". $\endgroup$ – AlexL Oct 4 '18 at 0:56
  • $\begingroup$ Great, thanks Alex! $\endgroup$ – Prince M Oct 4 '18 at 0:59
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Q1: The intent seems to be $Y$ is a closed subset of $X$ - by my reading, the possessive is attempting to clarify this. Perhaps it is a slight translation error? I am not good with French, so I cannot offer much perspective here.

Uncertainty 1: I think this is meant as "on the other hand", not with any specific mathematical meaning.

Uncertainty 2: The goal is to use the gluing axiom to come up with a continuous section which equals the section $s$ when restricted to $Y$. We'll use the open cover consisting of $X\setminus Y$ and an open set containing $Y$ - we will produce a continuous section on each of these, then show the sections agree on overlaps to give a section of $\mathcal{F}$ on $X$.

The section we choose on $X\setminus Y$ is the zero section. Easy enough.

In order to produce the open set containing $Y$ we proceed as follows. For each point $y\in Y$, there exists a neighborhood $U_y\subset Y$, a neighborhood $V_y\subset X$ so that $V_y=Y\cap U_y$ and a section $s'$ in $\Gamma(\mathcal{F},V_y)$ so that under the homomorphism we're investigating, the image of $s'$ agrees with $s$ (we can find these due to the definition of the induced sheaf on $Y$). The claim is that the $s'$ actually patch together to a section on the union of the $U_y$ - this is true because at any point $x$ in any intersection of the $U_y$, these sections agree: if $x$ is in $Y$, then the sections agree because they all agree with $s$, and if $x$ is not in $Y$, then each section is $0$ by the definition of being concentrated on $Y$. So all of these $s'$ patch together to form a section on $\bigcup U_y$ which equals $s$ on $Y$ and zero elsewhere.

Now we have covered $X$ by $X\setminus Y$ and $\bigcup U_y$, and we have sections of $\mathcal{F}$ on both which agree on the overlap. Thus these sections glue to form a global section, and by construction, it's image under the homomorphism in question must be $s$.

Uncertainty 3: Yes, this result in this generality really depends on $Y$ being closed. If $Y$ were open and not closed, we could find a point $x\in X$ so that $x\in \overline{Y}$ but not $x\in Y$. On the one hand, every section must have value $0$ here, but any continuous section's values here are also determined by their behavior in $Y$: this sets up some strange and undesirable conflicts that can lead to contradictions or only being able to consider the zero sheaf or something like that.

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  • $\begingroup$ Is the $U_y$ and $V_y$ backwards? $\endgroup$ – Prince M Oct 4 '18 at 1:17
  • $\begingroup$ The part I am really struggling with is where Serres set $V$ comes from, but I don't see what it would have to do with definition of induced sheaf since thats just $\pi^{-1}(Y)$ as a sheaf over $Y$? $\endgroup$ – Prince M Oct 4 '18 at 1:56
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The set $V$ that Serre asserts can be constructed as follows. Take $U \cap Y$ then view both $s$ and $s'$ as section into $\mathscr{F}(Y)$. They both still agree on the point $y$ so by the results of $n^{\circ}2$ they agree on some open neighborhood in $Y$ containing $y$, call it $W$. By definition of open in $Y$ we have $W = Y \cap O$ for some $O$ open in $X$. Then $V = O \cap U$ has the properties desired, i.e $V \subset U$ and both sections agree on $V \cap Y$.

With the existence of this $V$ established, one can combine this with the answer provided by KResier, which illuminates the method of Serres argument.

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