Show that $f^{-1}$ is continuous

Question

Let $E$ and $F$ two normed vector spaces, $A \subset E$ compact, $B \subset F$ and $f: A \to B$ is a bijective continuous function. As $f$ is bijective, we can defining the inverse function $f^{-1} : B \to A$ by $$f^{-1}(y)=x \iff f(x)=y.$$ Show that $f^{-1}$ is continuous.

Theorem : $f$ is continuous $\iff$ for each open set $V$ in $B$, $f^{-1}(V)$ is open in $A$.

I think is preferable to use this theorem instead of the sequentially continuous theorem.

I am stuck on this problem. Is anyone is able to help me a bit to continue this question?

http://math.stackexchange.com/questions/541082/proving-the-inverse-of-a-continuous-function-is-also-continuous — parsiad, Feb 06 '16 at 16:23
@par I know this is a similar question, but my question is more specific because $A \subset E$ and $B \subset F$, so the topology is a bit different. — Taj Mohamed Bandalandabad, Feb 06 '16 at 16:28

score 3 · Accepted Answer · answered Feb 06 '16 at 16:22

This is just a special case of a general fact about topology: If $X$ is compact, $Y$ is Hausdorff, and $f:X\to Y$ is a continuous bijection then $f^{-1}$ is continuous:

Since $f$ is a bijection, the inverse image of $S$ under $f^{-1}$ is just $f(S)$. So you have to show that $f(S)$ is open for every open set $S\subset X$. Since $f$ is a bijection this is the same as showing that $f(C)$ is closed for every closed set $C\subset X$. But $X$ compact, $C$ closed implies $C$ is compact, so $f(C)$ is compact, and now $Y$ Hausdorff implies $f(C)$ is closed.

score 1 · Answer 2 · answered Feb 06 '16 at 16:23

1

Hint: show that $f(C)$ is closed for every $C$ closed in $A$.

answered Feb 06 '16 at 16:23

Eduardo Longa

5,226

Show that $f^{-1}$ is continuous

2 Answers2