21
$\begingroup$

In a rather obscure article, I found (without proof) the following statement:

If $M$ is a closed orientable manifold, every degree $1$ map $f: M \rightarrow M$ is a homotopy equivalence.

Is this really true?

Using Poincare duality, it is easy to see that $f$ is a homology equivalence. But has $f$ to induce an isomorphism on $\pi_1$? Another (maybe related) result is Hopf's theorem: The degree classifies maps $M \rightarrow S^n$ up to homotopy equivalence.

(I am sorry if this question is too basic. Feel free to delete it in this case.)

$\endgroup$
2
  • 8
    $\begingroup$ It is in fact Problem 5.26 of Kirby's problem list (math.berkeley.edu/~kirby/problems.ps.gz), and still open as far as I know. $\endgroup$ – skupers Jan 23 '15 at 6:40
  • 1
    $\begingroup$ I don't know what dimension you're interested in, but perhaps it's worth pointing out that 3-manifold groups are known to be Hopfian. The upshot is that the assertion is true for irreducible 3-manifolds with infinite fundamental group, and perhaps also in greater generality than that... $\endgroup$ – HJRW Jan 27 '15 at 9:45
23
$\begingroup$

I believe that this is an open question in general, and the assertion is an old conjecture of Hopf. Some special cases were considered by Jean-Claude Hausmann, Geometric Hopfian and non-Hopfian situations. Geometry and topology (Athens, Ga., 1985), 157–166, Lecture Notes in Pure and Appl. Math., 105, Dekker, New York, 1987. I don't think there's been much progress since then, at least not that I could find via Mathscinet or Google Scholar.

It is easy to see that $f$ induces a surjection on $\pi_1$; if not, then $f$ factors through a non-trivial covering space of $M$, contradicting the degree-$1$ assumption. So if $\pi_1$ is Hopfian (any surjection from G to G is an isomorphism) then you get that $f_*$ is an isomorphism on $\pi_1$. There are, however, some non-Hopfian groups. Even when you know that $f_*$ is an isomorphism, you need more to show $f$ to be a homotopy equivalence; you'd need $f$ to induces homology isomorphisms with coefficients in $Z[\pi_1]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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