Proving that $f^{-1}$ is continuous 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?
 A: 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 $X$.  
A: Let $X,Y$ be topological spaces and let $g: X\to Y$ be a function. 
In general for $A\subseteq Y$ we have:$$g^{-1}(A^{\complement})=g^{-1}(A)^{\complement}$$

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.

So you and your textbook are saying equivalent things.
