# Proof that continuous function has continuous inverse

Assume that $A\subseteq\mathbb{R}^n$ is compact and that $\mathbf{f}:A\to\mathbb{R}^m$ is continuous and injective. Let $B=\mathbf{f}(A)\subseteq\mathbb{R}^m$ and show that $\mathbf{f}^{-1}(B)\rightarrow A$ is continuous.

I want to use the sequential definition of continuity.

Let $\mathbf{y} \in B$ and $(\mathbf{y}_n) \rightarrow \mathbf{y}$ with $\mathbf{y}_n \in B$ for each $n\in \mathbb{N}$.

Then since $B=\mathbf{f}(A)$, there are $\mathbf{x} , \mathbf{x}_n \in A$ such that $\mathbf{f}(\mathbf{x}) =\mathbf{y}$ and $\mathbf{f}(\mathbf{x}_n) = \mathbf{y}_n$ for each $n$.

Since $A$ is compact, it is closed and bounded. So $(\mathbf{x}_n)$ is bounded. We show that $(\mathbf{x}_n)$ converges.

It's enough to show that any convergent subsequence of $(\mathbf{x}_n)$ converges to the same point in $\mathbb{R}^n$. So let $(\mathbf{x}_{n_k})$ be a convergent subsequence that converges to $\mathbf{a}$. Since $A$ is closed, $\mathbf{a} \in A$. By continuity, $\mathbf{f}(\mathbf{x}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. Therefore, $(\mathbf{y}_{n_k})\rightarrow \mathbf{f}(\mathbf{a})$. But since $(\mathbf{y}_{n_k})$ is a subsequence of $(\mathbf{y}_n)$, we get that $\mathbf{y} =\mathbf{f}(\mathbf{a})$. Thus, $\mathbf{f}^{-1}(\mathbf{y})=\mathbf{a} =\lim_{k\to \infty} \mathbf{x}_{n_k}$. And every convergent subsequence of $(\mathbf{x}_{n})$ converges to $\mathbf{f}^{-1}(\mathbf{y}):=\mathbf{a}$, so that $(\mathbf{x}_n)\rightarrow \mathbf{a}$.

EDITED QUESTION: Does this proof work? I'm a bit concerned that I haven't used the fact that $\mathbf{f}({A})$ is a compact set. Overall, I'm not sure if I made a mistake somewhere.

• Do you have to show it with sequences? It is much easier in a general topology setting. – Jakob Elias Apr 25 '17 at 22:37
• @JakobElias Yes, with sequences. – CuriousKid7 Apr 25 '17 at 22:37
• It is a long time I did any general topology, so correct me if I am wrong. A continuous function from $R^n$ to $R^m$ maps compact set to compact set (to see this, consider the sub sequence criterion, a set is compact iff. every sequence contains converging sub-sequence). Now a closed subset of compact is compact, and a compact subset in $R^m$ is closed. In a nutshell, $f$ maps closed sets to closed sets. – Ran Wang Apr 25 '17 at 22:50

Start from your sequence $(\mathbf{y_n})\to \mathbf{y}$ and consider an arbitrary subsequence $(\mathbf{y_{n_k}})$ of $(\mathbf{y_n})$. By what you proved, you can find a subsequence $(\mathbf{y_{n_{k_j}}})$ of $(\mathbf{y_{n_k}})$ such that $(f^{-1}(\mathbf{y_{n_{k_j}}}))\to f^{-1}(\mathbf{y})$. This implies that the entire sequence $(f^{-1}(\mathbf{y_{n}}))\to f^{-1}(\mathbf{y})$ since otherwise, you could find a subsequence $(\mathbf{y_{n_k}})$ such that $\vert f^{-1}(\mathbf{y_{n_{k}}})-f^{-1}(\mathbf{y})\vert\ge \varepsilon>0$ for all $k$, which would be a contradiction. I am using this fact theorem on subsequences Does this answer your question?
• Ok, to be clear: I have already proven a theorem beforehand that says that if every convergent subsequence of a bounded sequence $(x_n)$ has the same limit $l$, then $(x_n)$ converges to $l$ as well. So I have already justified the "it's enough..." part. Having said this, do you see any issue with my proof as it stands now? I never used the fact that $f(A)$ is compact and want to be sure that my proof is valid anyway. – CuriousKid7 Apr 26 '17 at 1:51
• yes, you do not need to use the fact that $f(A)$ is compact since you start with a sequence which is already convergent in $f(A)$ – Gio67 Apr 26 '17 at 1:55
1. Since $f$ is continuous, $f(A)$ is compact and therefore closed.
2. Let $y_{n}\in f(A)$ a convergent sequence. By 1., we can supose $y_{n}=f(x_{n})\to f(x)=y$.
3. We want to show that $f^{-1}(y_{n})\to f^{-1}(y)$, i.e. $x_{n}\to x$. If $x_{n}\not\to x$, then there exists a subsequence such that $|x_{n_{k}}-x|\geq \epsilon$. Then we can extract a convergent subsequence $x_{n_{k_{j}}}$ of $x_{n_{k}}$ to, say, $x'$. But then we would have $f(x')\leftarrow f(x_{n_{k_{j}}})=y_{n_{k_{j}}}\to y=f(x)$. So, $f(x)=f(x')$ and then $x=x'$, a contradiction.