In an exercise I'm asked to prove the following:
Prove that every local homeomorphism is a continuous mapping
I was able to write a proof for this, but it turned out a title convoluted and I would like to know if there's any way of proving this more directly. This is what I did:
My proof
Let $f:(X,\tau) \to (Y,\pi)$ be a local homeomorphism.
Let $A \in \pi$. We want to prove that $f^{-1}(A) \in \tau$.
Let $K_x \in \tau$ be a neighbourhood of $x \in X$ such that $f(K_x) \in \pi$ and $f|_{K_x}:K_x \to f(K_x)$ is an homeomorphism.
So let's first divide the set $A$ into small pieces: So let's consider the sets: $A \cap f(K_x) \in Y$ for every $x \in X$. Now, let's send each piece back to the set $X$ using the function $f|_{K_x}$: $f|_{K_x}^{-1} (A \cap f(K_x))$, and now that we have each peace in the set $X$, let's put them all together and we must get $f^{-1}(A)$:
$$f^{-1}(A) = \bigcup _{x \in X} f|_{K_x}^{-1} (A \cap f(K_x)) $$
I proved that these sets are the same, But I'm not going to put the proof here so the post doesn't become too long.
Now we know that $A \cap f(K_x)$ is open in $f(K_x) \subseteq Y$. This means that $f|_{K_x}^{-1} (A \cap f(K_x))$ is open in $K_x \subseteq X$. So $\exists B_x \in \tau: f|_{K_x}^{-1} (A \cap f(K_x)) = \underbrace{B_x \cap K_x}_{\in \tau}$. So we have the following:
$$f^{-1}(A) = \bigcup _{x \in X} \underbrace{f|_{K_x}^{-1} (A \cap f(K_x))}_{\in \tau}$$
So, because $f^{-1}(A)$ is the union of open sets, it's open, making $f$ a continuous function.
Is my proof correct? Is there a more straightforward way of proving this? Thanks for the help
It looks correct, but I would do it as follows. Take $x\in X$; you want to prove that $f$ is continuous at $X$. Let $V$ be a neighborhood of $f(x)$ then; you want to prov that $f^{-1}(V)$ is a neighborhood of $x$. Pick a neighborhood $K$ of $x$ such that $f|_K$ is a homeomorphism from $K$ onto $f(K)$. Then, since $V\cap f(K)$ is a neighborhood of $f(x)$ in $f(K)$, $f^{-1}\bigl(V\cap f(K)\bigr)$ is a neighborhood of $x$. But then, since $f^{-1}\bigl(V\cap f(K)\bigr)\subset f^{-1}(V)$, $f^{-1}(V)$ is a neighborhood of $x$ too.