It's easy to show using Stokes theorem that a compact orientable manifold with boundary cannot retract to its boundary, by choosing a volume form. But for the non-orientable case I don't know if this is true. Is there a non-orientable manifold with a retraction of to its boundary?

Thanks in advance.


2 Answers 2


Here's a proof which worked for both orientable and non-orientable manifolds. The proof comes from (my memory of) Milnor's book "Topology from the Differentiable Viewpoint". The proof only shows there is no smooth retraction. (With more work, one can show that there is a continuous retraction iff there is a smooth one, so this can be turned into a proof of the general statement.)

Suppose for a contradiction that $r:M\rightarrow \partial M$ is a retraction of a compact manifold $M$ onto its boundary. By Sard's theorem, there is a regular value $p\in \partial M$.

Then $r^{-1}(p)$ is a $1$-d submanifold in $M$. It is closed, being the inverse image of the closed set $\{p\}$, so is compact. A compact $1$-d manifold is homeomorphic to a disjoint union of a finite number of circles and closed, bounded intervals. In particular, the boundary of $r^{-1}(p)$ has even cardinality.

Now, the key point is that the boundary of $r^{-1}(p)$ must lie in $\partial M$. But $r^{-1}(p)\cap \partial M = \{p\}$ (since $r$ is a retraction), so is not of even cardinality.

  • $\begingroup$ Wow. I am so impressed by this proof... $\endgroup$
    – Vadim
    Oct 10, 2017 at 22:18

No, this is never the case. The simplest way to see this is to consider the homology of the manifold and its boundary. Using $\mathbb{Z/2Z}$ coefficients, we can avoid questions of orientability.

For the retraction to exist, we would need to have the composition:$$H_{n-1} (\partial M; \mathbb{Z/2Z}) \rightarrow H_{n-1} ( M; \mathbb{Z/2Z}) \rightarrow H_{n-1} (\partial M; \mathbb{Z/2Z})$$

be the identity. I will leave it as an exercise to you to demonstrate that this cannot occur. Consider what happens to stuff on the boundary when you stick it into the manifold at large.

  • $\begingroup$ Thanks for answer. Why considering $\mathbb{Z}_2$ kills the problem of orientability? And I could not understand your hint. "...into the manifold at large"? $\endgroup$
    – user40276
    Jul 27, 2014 at 23:46
  • $\begingroup$ Non orientability means you will have sign problem after you triangulate it, and this will make the top homology with integer coefficients to be zero. However, with Z_2 coefficients the sign will not be a problem. $\endgroup$ Jul 28, 2014 at 0:45

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