$M,N$ metric spaces, $\phi:M\to N$ a surjective open map. Show that the map $f:N\to P$ is continuous iff $f\circ \phi$ is continuous

Question

I need to show the following:

$M,N$ metric spaces, $\phi:M\to N$ a surjective open map. Show that the map $f:N\to P$ is continuous iff $f\circ \phi$ is continuous

In order to show that the composite is continuous, I need to show that its inverse is an open map. I have found this question but it's incomplete since it doesn't assume surjectivity.

I'd also like someintuition on why this is useful. I appreciate any help <3

Answer 1

Suppose $f\circ\phi$ is continuous. We need to prove $f$ is continuous. Let $U\subseteq P$ be open. We need to show that $f^{-1}(U)$ is open. Since $f\circ\phi$ is continuous, $(f\circ\phi)^{-1}(U)=\phi^{-1}( f^{-1}(U))$ is open in $M$. Since $\phi$ is surective, we have $\phi(\phi^{-1}(f^{-1}(U)))=f^{-1}(U)$ and since $\phi$ is open $f^{-1}(U)$ is open. We are done.

Note that the converse is not necessarily true -

Let $M=\mathbb R$ and $N=P=\{-1,0,1\}$ with the induced metric from $\mathbb R$. Define $\phi:\mathbb R\to N$ by $$\phi(x)=\left\{\begin{array}[rcl] &\dfrac{x}{|x|}& if & x\neq0\\0& if&x=0 \end{array}\right.$$ Then $\phi$ is open and surjective (it is not continuous as in the question you linked)

Let $f:N\to P$ be the identity map. Then $f$ is continuous but $f\circ\phi=\phi$ is not.