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I'm self-studying topological manifold from the begining and I have many maybe trivial questions;

I would like to show that every homeomorphism is a local homeomorphism;

A continuous map $f:X\rightarrow Y $ is said a Local Homeomorphism if every point $x\in X$ has a neighborhood $U\subset X$ such that $f(U)$ is an open subset of $Y$ and $f|_{U}:U\rightarrow f(U)$ is an homeomorphism.

I have proceeded in this way:

If $f$ is homeomorphism,$f^{-1}$ is continuous,hence for each $x \in X$ all its open neighborhoods $U_x$ are such that $f(U_x)$ is open in $Y$.

Now I need to show that $\forall x\in X$ there is an open neighbothood $U_x$ such that: $$f|_{U_x}: U_x \rightarrow f(U_x) $$$$f^{-1}|_{U_x}: f(U_x) \rightarrow U_x $$ are both continuous (in other words $f|_{U_x}$ is an homeomorphism )

I know that if a function $g:M\rightarrow N $ is continuous then every restriction $g|_A$ with $A$ open subset of $M$ is continuous; hence $\forall x\in X$ I have that $f|_{U_x}: U_x \rightarrow f(U_x) $ is continuous in every open neighbothood $U_x$ of $x \in X$; moreover $f(U_x)$ is an open subset of $Y$ and $f^{-1}$ is continuous, then $f^{-1}|_{U_x}: f(U_x) \rightarrow U_x $ is continuous too.

Could someone check if the proof is correct?

Thank you.

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    $\begingroup$ Looks OK at a glance. However, you don't need to do all this work: say $f : X\to Y$ is your homeomorphism. $X$ is a neighborhood of any point $x\in X$, $f(X) = Y$ is open in $Y$, and $\left. f\right|_X = f$ is a homeomorphism. $\endgroup$ – Stahl May 20 '17 at 8:30
  • $\begingroup$ @Stahl thank you a lot. Your proof is much better than mine $\endgroup$ – Simone P May 20 '17 at 8:36
  • $\begingroup$ If you're happy with this, I'll post it as an answer so the question doesn't appear unanswered. $\endgroup$ – Stahl May 20 '17 at 8:36
  • $\begingroup$ ok...I'm satisfied. $\endgroup$ – Simone P May 20 '17 at 8:39
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Your proof looks OK at a glance. One note: you might want to make a remark about $f$ being bijective so that $f^{-1}$ actually defines a function, and then $\left.f^{-1}\right|_U$ does as well for any $U\subseteq Y$ (as $\left.f\right|_V$ will be a bijection for any $V\subseteq X$). After all, a homeomorphism is first and foremost a bijection, and you never mention/address this aspect explicitly, although you use it implicitly in your treatment of $f^{-1}$ as a function.

However, you can give a much cleaner proof: say $f:X\to Y$ is your homeomorphism. $X$ is a neighborhood of any point $x\in X$, $f(X)=Y$ is open in $Y$, and $\left.f\right|_X=f$ is a homeomorphism.

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