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I've tried prove the follow statement

Let $K \subset \mathbb{R^n}$ a compact set and $f: K \rightarrow \mathbb{R^m}$ a continuous map. If $f$ is injective then its inverse $g:f(K)\rightarrow K$ is continuous

My proof:

Let $\{y_n\}_{n \in \mathbb{N}} \subset f(K)$ such that $y_n = f(x_n), \ x_n \in K$ and $y_n \rightarrow y \in f(K)$. We have a subsequence $\{x_{n_k}\}_{k \in \mathbb{N}}$ of $\{x_{n}\}_{n \in \mathbb{N}}$ such that $x_{n_k} \rightarrow x \in K \Rightarrow y_{n_k} = f(x_{n_k}) \rightarrow f(x)$, because $f$ is continuos, also we have $f(x) = y$ because $y_n \rightarrow y \Rightarrow y_{n_k} \rightarrow y $.

Note $g(y_{n_k}) = g(f(x_{n_k})) = x_{n_k} \rightarrow x = g(f(x)) = g(y)$.

Then g is continuous. [QED]

For me, it looks good but I'm not 100% sure of that prove it's correct, if somebody could give a feedback I would appreciate it.

Thank you.

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  • $\begingroup$ More generally (1). If $K$ is a compact space and $f:K\to L$ is continuous then $f(K)$ is a compact sub-space of $L$. ... (2). If $K$ is a compact Hausdorff space and $L$ is a Hausdorff space and if $f:K\to L$ is a continuous injection then $f:K\to f(K)$ is a homeomorphism. That is, $f^{-1}:f(K)\to K$ is continuous. $\endgroup$ – DanielWainfleet Apr 8 '18 at 7:29
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Technically what you have proved is that for every $y_n\to y$ there is a subsequence $y_{n_k}$ such that $g(y_{n_k})\to g(y)$. That is not exactly what you needed (you would need $g(y_n)\to g(y)).$


A cleaner way would be:

Let $C$ be a closed set in $K$. Then $g^{-1}(C)=f(C)$.

Now $C$ is closed in a compact subset of $\Bbb R^n$ so $C$ is compact itself, therefore $f(C)$ is compact since $f$ is continuous, hence $g^{-1}(C)=f(C)$ is closed.

We proved that for every closed set $C$, $g^{-1}(C)$ is closed, so $g$ is continuous.

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