# Continuous map to quotient space

Let $X,Y,Z$ be topological spaces, and suppose $\pi:Y\rightarrow Z$ is a quotient map. Is a continuous map $f:X\rightarrow Z$ necessarily the composition of a continuous map $g:X\rightarrow Y$ with $\pi$? What if $X,Y,Z$ are smooth manifolds and we replace 'continuous' with 'smooth'?

-
If $X=Z$ and $f=\mathrm{id}$, the condition would be equivalent to $Z$ being a retract of $Y$. –  Arturo Magidin Apr 19 '12 at 3:31

No. Let $X$ and $Z$ be circles and $Y=\mathbb{R}$. There is a continuous (and in fact smooth) bijective map $X \to Z$, but it won't factor through $Y$, a map from a circle to the real line cannot be injective.
Definitively not. Take a Lie groups $G$ and $H$ and think that if we can factor $G\to H$ through $G\to G/N$ for some normal subgroup $N$ then $\ker(G\to H)\subseteq N$.