Let $f\colon X\to Y$ be a surjective continuous map between two topological spaces such that $X,Y$ are path-connected and such that every fibre $f^{-1}(y)$ is connected, for each $y\in Y$. Is there always a continuous section $s\colon Y\to X$ (i.e. a continuous map $s$ such that $f\circ s$ is the identity on $X$)? If no, what kind of sufficient conditions could we impose on $f$ ? For example, is it better if $Y$ is supposed to be simply connected? Or if we choose $Y$ to be the interval $[0,1]$ ?
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asked Nov 21, 2015 at 23:02
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3$\begingroup$ The hairy ball theorem says that there is no section to the unit sphere bundle over the $2$-sphere : $US^2\to S^2$. $\endgroup$– Olivier BégassatCommented Nov 21, 2015 at 23:12
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1$\begingroup$ Thanks for the nice comment. I did not know about this result. Are there some examples where $Y=[0,1]$ ? $\endgroup$– Jérémy BlancCommented Nov 21, 2015 at 23:16
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$\begingroup$ You could consider $f^{-1}$ as a multivalued function from $Y$ to $X$ and then look at the various "continuous-selection-theorems" out there. For instance, if $f:[0,1] \to [0,1]$ is an open suryection and $f^{-1}(y)$ is compact for each $y$, then you have a continuous section (we don´t need to assume that $f$ is continuous or that $f^{-1}(y)$ is connected). $\endgroup$– Ramiro de la VegaCommented Nov 25, 2015 at 19:22
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2 Answers
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No, even if $Y=[0,1]$. The piecewise linear continuous nondecreasing surjection $f:[0,1] \rightarrow [0,1]$ which maps $[1/3,2/3]$ to $1/2$ and is otherwise 1-1 and linear has no continuous section.
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Asking when a continuous map $f : X \to Y$ has a continuous section is analogous to asking when a Diophantine equation over $\mathbb{Z}$ has a solution over $\mathbb{Z}$; see, for example, this blog post. This suggests that in general the problem will be hard and that there will be many obstructions, both local and global, and both of these are true.
For example, asking that $f$ be surjective, or equivalently that the fibers $f^{-1}(y)$ be nonempty, is the analogue of asking that the Diophantine equation have a solution $\bmod p$ for all primes. The next most local thing to ask for is that around each $y \in Y$ there is a neighborhood $U$ on which $f$ has sections; this is loosely the analogue of asking that the Diophantine equation have $p$-adic solutions for all primes. This is true, for example, for all covering maps, but of course there is a global obstruction to having a section in this case, namely that the induced map on $\pi_1$ must also have a section.
Sufficient conditions are hard to come by, but I can list a bunch of necessary conditions. The problem becomes much more amenable to homotopy theory if $f$ is a fibration; then all of the obstructions are global and homotopy-theoretic in nature and it is possible in principle to completely describe them.
answered Nov 22, 2015 at 6:32