# Is every degree 1 self-map a homotopy equivalence?

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.)

• 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. Jan 23, 2015 at 6:40
• 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...
– HJRW
Jan 27, 2015 at 9:45

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 $Z[\pi_1]$.