# Cancellation of contractibility of fibres

Suppose given maps $f:X \to Y$ and $g:Y \to Z$ such that $f$ and $g \circ f$ both have contractible fibres. Then does $g$ have contractible fibres?

And, the same question, but with the maps assumed to be morphisms of algebraic varieties and contractible replaced with isomorphic to $\mathbb{C}^n$ (for varying $n$).

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Without an additional assumption about properness or a fibration-like property, there are pretty simple counterexamples: $X=[0,1)$, $Y=S^1$, $Z=\ast$, $f(x)=e^{2 \pi i x}$.

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If your spaces are path-connected, locally compact separable metric spaces (with some additional local connectedness assumptions), and the maps are all proper (inverse images of compact sets are compact) and onto, then the Vietoris mapping theorem can be applied.

You can conclude that the map $g$ is weak homotopy equivalence. If $g$ is a fibration, this implies that its fibres are all weakly contractible. If $Y$ and $Z$ are CW-complexes, then the fibres of $g$ are of the homotopy type of a CW complex (by a theorem of Milnor) and hence are contractible.

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I assume here that "fibres" means "inverse images of points". I hope I'm not misinterpreting.

If $X$ is a point, then all of the fibres of $f$ and $g\circ f$ are either singletons or empty, hence contractible. But the map $g$ can be anything at all.

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The empty space is not contractible - it is not homotopy equivalent to a point. In particular, this implies that both $f$ and $g\circ f$ are surjective, so the fibres of $g$ are non-empty. This is a loong way from knowing they are contractible though. –  David Roberts Mar 9 '11 at 6:54
Ah. Gotcha. Thanks. –  Steven Landsburg Mar 9 '11 at 8:10