# Inducing a sheaf on an open subset, or any subset?

I am working out of Serre's FAC - if my definition of a sheaf is bizarre to you, that is because it is the older definition. In modern language this object is typically referred to as etale space (I can't figure out how to do the accents). Most authors at least briefly mention the categorical equivalence between the two.

Serre Defines:

Let $X$ be a topological space. A sheaf of abelian groups on $X$ (or simply a sheaf ) consists of:

(a) A function $x \to \mathscr{F}_x$, giving for all $x \in X$ an abelian group $\mathscr{F}_x$,

(b) A topology on the set $\mathscr{F}$, the sum of the sets $\mathscr{F}_x$.

If $f$ is an element of $\mathscr{F}_x$, we put $\pi(f) = x$; we call the mapping of $\pi$ the projection of $\mathscr{F}$ onto $X$; the family in $\mathscr{F} \times \mathscr{F}$ consisting of pairs $(f,g)$ such that $\pi(f) = \pi(g)$ is denoted by $\mathscr{F}+\mathscr{F}$.

Having stated the above definitions, we impose two axioms on the data (a) and (b):

(I) For all $f \in \mathscr{F}$ there exist open neighborhoods $V$ of $f$ and $U$ of $\pi(f)$ such that the restriction of $\pi$ to $V$ is a homeomorphism of $V$ and $U$.(In other words, is a local homeomorphism).

(II) The mapping $f \mapsto -f$ is a continuous mapping from $\mathscr{F}$ to $\mathscr{F}$, and the mapping $(f, g) \mapsto f + g$ is a continuous mapping from $\mathscr{F}+\mathscr{F}$ to $\mathscr{F}$.

Serre asserts in

Chapter 1: Sheaves, Section 1: Operations on sheaves, Subsection 4: Glueing sheaves,

Let $\mathscr{F}$ be a sheaf on $X$, and let $U$ be a subset of $X$; the set $\pi^{-1}(U) \subset \mathscr{F}$, with the topology induced from $\mathscr{F}$, forms a sheaf over $U$, called a sheaf $\textit{induced}$ by $\mathscr{F}$ on $U$.

My question is about additional properties of the subset $U$, and I only pursue this because Serre is typically very careful about stating assertions precisely. I glossed through a proof of this, but upon returning to it I realize my proof requires heavily that $U$ be open in $X$, and Serre did not state $U$ open?

One can see plainly that $U$ cannot be empty for axiom (a), but this is a trivial matter. Edit: $U$ can be empty. Whoops! It does seem, however, you could lose axioms (I) and (II) if $U$ is not open but I don't have a counter example. It may be beneficial for me to come up with a counter example on my own, so please feel free to recommend an explicit sheaf where one could be found, or steer me in the right direction. I am somewhat new to sheaves.

• Why can't $U$ be empty? If $U$ is empty, the empty function assigns an abelian group to every element of $U$ (vacuously). In fact, you can restrict a sheaf $\mathcal{F}$ to an arbitrary subspace $Y\subseteq X$, as Serre writes. This is the inverse image sheaf $i^{-1}\mathcal{F}$ along the inclusion $i\colon Y\to X$: en.wikipedia.org/wiki/Inverse_image_functor – Alex Kruckman Aug 16 '18 at 21:12
• Yes of course, I don’t know what I was thinking regarding the empty function. – Prince M Aug 16 '18 at 21:18
• And ok, so Serre asserts it the way it is! I will just try and prove it via the definition. Since I am aware of the equivalence of categories, I could extend the results from the link you provide, but it could help my intuition to prove it straight from Serres definition. – Prince M Aug 16 '18 at 21:41
• I am so far unable to verify the local homeomorphism. I am taking the $V$ that exists by axiom (I) and then looking at $V \cap \pi^{-1}(U)$ for a homeomorphism onto the image, but why should this be a homeomorphism? – Prince M Aug 17 '18 at 0:16
• Ok, I wrote an answer. – Alex Kruckman Aug 17 '18 at 1:01

Suppose $Y$ is any subset of $X$, equipped with the subspace topology (I'm going to use $Y$ instead of $U$, since there's a $U$ mentioned in axiom (I)). We also equip $\mathcal{F}' = \pi^{-1}(Y) \subseteq \mathcal{F}$ with the subspace topology.
For (I), suppose $f\in \mathcal{F}'$, so $\pi(f)\in Y$. Then there is an open neighborhood $V$ of $f$ in $\mathcal{F}$ and an open neighborhood $U$ of $\pi(f)$ in $X$ such that $\pi$ restricts to a homeomorphism $V\to U$. Then $U' = U\cap Y$ is an open neighborhood of $\pi(f)$ in $Y$, and $V' = V\cap \mathcal{F}'$ is an open neighborhood of $f$ in $\mathcal{F}'$. Now check that $V' = \pi^{-1}(U')$. Since $\pi$ restricts to a homeomorphism $V\to U$, it restricts further to a homeomorphism $V' = \pi^{-1}(U')\to U'$.
Similarly, for (II), the continuous map $f\mapsto -f$ restricts to a continuous map $\mathcal{F}'\to \mathcal{F}'$, and the continuous map $(f,g)\mapsto f+g$ restricts to a continuous map $\mathcal{F}'+\mathcal{F}'\to \mathcal{F}'$.