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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'?

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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
up vote 5 down vote accepted

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.

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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 $\ker(G\to H)\subseteq N$.

EDIT: Of course, this is incorrect because I was thinking in Lie groups--the result I stated is true if you want a Lie group map to be the factor map. The basic idea is that this isn't even true for sets. The map has to respect equivalency classes.

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