Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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'?

share|improve this question
1  
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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.