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

Question

We are given $f:X\to Y$ to be a bijective continuous function and my textbook has the following statement:

We shall prove that images of closed sets of $X$ under $f$ are closed in $Y$; this will prove continuity of the map $f^{-1}$.

I think the above statement is incorrect.

$f^{-1}:Y\to X$

The condition that $f^{-1}$ be continuous says that for each open set $U$ of $X$, the inverse image of $U$ under the map $f^{-1}$ is open in $Y$. But the inverse image of $U$ under the map $f^{-1}$ is the same as the image of $U$ under the map $f$. So, for $f^{-1}$ to be continuous, for each open set $U$ of $X$, $f(U)$ is open in $Y$. My textbook says "closed" instead of "open". Who is right?

Answer 1

Recall that a function $f:X\to Y$ (where $X,Y$ are metric spaces) is continuous if and only if $f^{-1}(G)$ is closed in $X$ whenever $G$ is closed in $Y$.

In your case $f^{-1}:Y\to X$ is continuous if and only if $(f^{-1})^{-1}(G)=f(G)$ is closed in $Y$ whenever $G$ is closed in $X$.

Answer 2

Let $X,Y$ be topological spaces and let $g: X\to Y$ be a function.

In general for $A\subseteq Y$ we have:$$g^{-1}(A^{\complement})=g^{-1}(A)^{\complement}$$

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This makes it possible to prove that the following statements are equivalent:

$1)$ For every closed $F\subseteq Y$ the set $g^{-1}(F)\subseteq X$ is closed.

$2)$ For every open $U\subseteq Y$ the set $g^{-1}(U)\subseteq X$ is open.

$(1\implies2)$

Let $U\subseteq Y$ be open

Then $U^{\complement}$ is closed so that $g^{-1}(U)^{\complement}=g^{-1}(U^{\complement})$ is closed.

That means exactly that $g^{-1}(U)$ is open.

$(2\implies1)$

Let $F\subseteq Y$ be closed.

Then $F^{\complement}$ is open so that $g^{-1}(F)^{\complement}=g^{-1}(F^{\complement})$ is open.

That means exactly that $g^{-1}(F)$ is closed.

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So you and your textbook are saying equivalent things.