I have just proved the result that if, $f:X \to Y$ is compact then for any $B\subset Y$ the map $g: f^{-1}(B) \to B$ is also proper.
So in particular we can say that the constant map,$f^{-1}(y) \to y$ is also proper.
Now i was trying to show that $f^{-1}(y)$ is also compact. I thought of using the result that $X$ is compact iff $p:X*Y \to Y$ is closed for all spaces $Y$ ( $p$ denotes the projection map and $X*Y$ is the product of two topological spaces $X$ and $Y$)
So i defined $p:f^{-1}(y)*Z \to Z$. Now how can i use the fact that the constant map,$f^{-1}(y) \to y$ is proper to show that $p$ is closed ?
The definition of proper maps that i am using is :
A map $f:X \to Y$ is said to be a proper map if $f*I:X*Z \to Y*Z$ is closed where $I:Z \to Z$ is the identity map.