Every local homeomorphism is an open map (Goldblatt's Topoi)

Question

This is at the top of page 97.

A sheaf is a bundle with some additional topological structure. Let $I$ be a topological space, with $\Theta$ its collection of open sets. A sheaf over $I$ is a pair $(A, p)$ where $A$ is a topological space $p : A \to I$ is a continuous map that is a local homeomorphism. This means that each point $x \in A$ has an open neighborhood $U$ in $A$ that is mapped homeomorphically by $p$ onto $p(U)$, and the latter is open in $I$. The category $\textbf{Top}(I)$ of sheaves over $I$ has such pairs $(A, p)$ a objects, and as arrows $k:(A, p) \to (B, q)$ the continuous maps $k : A \to B$ such that $q\circ k = p$ commutes. Such $k$ is in fact an open map (as is a local homeomorphism) and in particular $\text{Im} \ k = k(A)$ will be an open subset of $B$.

Here is my proof attempt.

A local homeomorphism $f: A \to I$ is an open map.

Proof. Let $U \subset A$ be open. Then for each $x \in U, \exists \ V_x \subset A$ which is open and such that $g = f|_{V_x} : V_x \simeq f(V_x)$ is a homeomorphism. Thus $V_x \cap U = $ an open set in $V_x$ and so $g(V_x \cap U) = g(V_x) \cap g(U)$ is open in $g(V_x)$.

Here I'm not sure whether unioning over $x$ would get us there.

Let's just focus on the local homeomorphism and I will make another question for the map $k$ if I need to.

Does this answer your question? Show that every local homeomorphism is continuous and open therefore bijective local homeomorphism is a homeomorphism — Arnaud D., Apr 18 '20 at 21:14
@ArnaudD. nope because of bijectivity added there. And I read the answers there, and still had the same question. — Daniel Donnelly, Apr 19 '20 at 16:55
The first part is to show that a local homeomorphism is open, without any assumption of bijectivity. — Arnaud D., Apr 19 '20 at 17:13
@ArnaudD. I am more happy with the answer below than any answer provided in that linked question. — Daniel Donnelly, Apr 19 '20 at 19:57

score 3 · Accepted Answer · answered Apr 18 '20 at 21:12

3

Let $y\in f(U)$. You want to find an open neighborhood of $y$ which is contained inside $f(U)$. Let $x\in U$ with $f(x) = y$. Using the local homeomorphism property, there is an open neighborhood $V$ of $x$ such that $f|V$ is a homeomorphism of $V$ onto $f(V)$. Then $f|V$ is an open map; since $V \cap U$ is an open neighborhood of $x$ in $V$, then $f(V \cap U)$ is an open neighborhood of $y$ which is contained in $f(U)$.

answered Apr 18 '20 at 21:12

kobe

41,901

so, since $f\vert_V$ is an open map because it is a homeomorphism from $V$ to $f(V)$, the image of $V\cap U$ is open because it's $f(U\cap V)=f\vert_V(U)$. Therefore $f(U\cap V)\subset f(U)$ and this shows that $f(U)$ is open because it is neghbourhood of each of its points. – Vajra Sep 30 '21 at 08:38
@Vajra yes, it looks like you got it. – kobe Sep 30 '21 at 15:26
I am a bit unsure about the last step. Let's call $(Y,\mathcal{T}{Y})$ the topology space on Y, and $(V,\mathcal{T}{V})$ the subspace on V induced by $\mathcal{T}{Y}$. According to my understanding, local homomorphism only implies that $f(U\bigcap V)$ is open in$(V,\mathcal{T}{V})$. But how do we know that an open set in $(V,\mathcal{T}{V})$ must be an open set in $(Y,\mathcal{T}{Y})$? – Jake ZHANG Shiyu Sep 17 '22 at 23:59
Oh, I think I got it because local homeomorphism maps an open neighborhood V in X onto an open set $f(V)$ in Y. So every open set in $(V,\mathcal{T}{V})$ is still open in $(Y,\mathcal{T}{Y})$. – Jake ZHANG Shiyu Sep 18 '22 at 00:18
@JakeZHANGShiyu that’s correct. – kobe Sep 18 '22 at 20:07

Every local homeomorphism is an open map (Goldblatt's Topoi)

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