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I am working on the following exercise from Lawson's Topology: A Geometric Approach:

Apply Invariance of Domain

(If $U$ is an open subset of $\mathbb{R}^n$ and $f:U\rightarrow\mathbb{R}^n$ is $1$-$1$ and continuous, then $f$ is an open map)

to prove that if $M$ and $N$ are $n$-manifolds and $f:M\rightarrow N$ is $1$-$1$ and continuous then $f$ is an open map.

I am really stuck. I can't see how to get a map from an open set in $\mathbb{R}^n$ to $\mathbb{R}^n$ involved to apply Invariance of Domain. Any help is greatly appreciated.

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Hint: You want to show that $f(U)$ is open in $N$ for all open $U$ in $M$. Note that as $f$ is continuous, for all $x\in U$, there are coordinate charts $U_x \subset U$ so that $f(U_x)$ lie in a coordinate chart $V_x$ of $N$. And you have

$$U = \bigcup_{x\in U} U_x$$

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  • $\begingroup$ Lawson doesn't mention "coordinate charts" so I am not sure what this means. I know that for each $x\in U$ there is a nbd $U_x$ that is homeomorphic to an open set in $\mathbb{R}^n$. Then $f(U)=f(\bigcup\limits_{x\in U} U_x)=\bigcup\limits_{x\in U}f(U_x)$. Now what can I say about the sets $f(U_x)$? $\endgroup$
    – 1234
    Jun 23 '16 at 15:56
  • $\begingroup$ @1234 : $f(U_x) \subset V_x$ and $V_x$ is also homeomophic to an open set in $\mathbb R^n$. $\endgroup$
    – user99914
    Jun 23 '16 at 15:58
  • $\begingroup$ I am guessing your $V_x$ is defined as: For each $x\in U$ there is a nbd $V_x$ of $f(x)$ that is homeomorphic to an open set in $\mathbb{R}^n$. But then how do we know that $f(U_x)\subset V_x$? $\endgroup$
    – 1234
    Jun 23 '16 at 16:10
  • $\begingroup$ @1234 $f$ is continuous, so $f^{-1}(V_x)$ is open, pick $U_x$ small so that $U_x \subset f^{-1}(V_x)$. $\endgroup$
    – user99914
    Jun 23 '16 at 16:57
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    $\begingroup$ Ah! I wanted to say this from the beginning but I couldn't think of how to make this precise. We can easily get a continuous 1-1 map between our open sets in $\mathbb{R}^n$ using f and the homeomorphisms we have. Then we can apply invariance of domain to get the result. Man, frustration really get's in the way of an easy solution! $\endgroup$
    – 1234
    Jun 24 '16 at 17:46

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