Follow up on another question I asked recently: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism
Definition: Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces. A
function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism if for
every point $x \in X$ there exists an open set $U \subseteq X$
containing $x$ and an open set $V \subseteq Y$ such that the restriction ${\displaystyle f|_{U}:U\to V\,}$ is a homeomorphism.
This definition is a bit alarming because it starts with "…if for
every point $x \in X$ there exists an open set $U \in \mathcal{T}$…", makes it seem like a property of the underlying space. Can we always find an open $U$? But anyways.
Objective: Show that every local homeomorphism is continuous and open
therefore bijective local homeomorphism is a homeomorphism
Proof:
(Honestly not sure what I am doing but proceed regardless)
Let $(X, \mathcal{T})$ and $(Y, \mathcal{J})$ be topological spaces and
function ${\displaystyle f:X\to Y\,}$ is a local homeomorphism. We will show that $f$ is continuous and open.
First show $f$ is continuous.
$f$ is continuous if for all $V \in \mathcal{J}, f^{-1}(V) \in \mathcal{T}$. Take some $V \in \mathcal{J}$, then $V$ is a subspace equipped with subspace topology $\mathcal{J}_V = \{V \cap W| W \in \mathcal{J}\}$.
Consider the inverse of the restriction $f^{-1}|_U$ on an open set in $\mathcal{J}_V$, then $f^{-1}|_U(V \cap W) = f^{-1}|_U(V) \cap f^{-1}|_U(W) $$= f^{-1}(V) \cap U \cap f^{-1}(W) \cap U = f^{-1}(V) \cap f^{-1}(W) \cap U$.
Then $f^{-1}(V) = f^{-1}(W) \cup U \cup f^{-1}|_U(V \cap W)$. We note all the sets on the right hand side are open. In particular, $U$ is open, $f^{-1}|_U(V \cap W)$ is open by definition of homeomorphism (?? $f^{-1}(W)$ ??), hence $f$ is continuous. ($\leftarrow$ something wrong here!)
Next show $f$ is open.
$f$ is open if $\forall U \in \mathcal{T}, f(U) \in \mathcal{J}$. Consider the restriction $f|_U$ on the subspace topology on $U$, $\mathcal{T}_U = \{U \cap M | M \in \mathcal{T}\}$. $f|_U(U \cap M) = f|_U(U) \cap f|_U(M) = V \cap f(M) \cap f(U)$
Then $f$ is open since $f(U) = f|_U(U \cap M) \cup V \cup f(M)$ and $f|_U(U \cap M)$ is open by definition of homeomorphism, $V$ is open in $\mathcal{T}$ (?? $\cup f(M)$ ??) ($\Leftarrow$ another mistake here)
I'm not quite sure how to proceed with showing bijective + continuous + open + local = homeomorphism part.
Can someone help me fix those two problems and give me some ideas how
to conclude that bijective local homeomorphisms are homeomorphisms?
Best Answer
Your attempts are, unfortunately, flawed.
Since you know about local properties of $f$, it is better showing that $f$ is continuous at each point.
Let $x\in X$; we want to show that, for every open neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ such that $f(U)\subseteq V$. Let $U_x$ be an open neighborhood of $x$ and $V_x$ an open set in $Y$ such that $f$ induces a homeomorphism $f_{U_x}\colon U_x\to V_x$ and choose any open neighborhood $V$ of $f(x)$.
Then $V\cap V_x$ is an open set in $Y$ containing $f(x)$,
Now you want to prove that $f$ is open. Let $A$ be open in $X$ and, for each $x\in A$, choose open sets $U_x\subseteq X$ and $V_x\subseteq Y$ so that $x\in U_x$ and $f$ induces a homeomorphism between $U_x$ and $V_x$.
For each $x\in A$, $f(U_x\cap A)$ is open in $V_x$, so it is open in $Y$ as well. Therefore $$ \bigcup_{x\in A}f(U_x\cap A) $$
If $f$ is bijective, then $f^{-1}$ exists and it is continuous