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A well-known theorem asserts that degree 1 maps induce surjections of the fundamental group. I am looking for a partial converse.

Is it true (under suitable assumptions) that a map between compact, aspherical manifolds of the same dimension has degree different from zero if the induced homomorphism between fundamental groups is surjective?

I am interested in maps between knot complements (mapping boundary to boundary) but any hint to the literature for some other special instances will be appreciated.

Michael Albanese
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user39082
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  • No, it's not true. The constant map $S^2 \to S^2$ is surjective on $\pi_1$ (because $\pi_1(S^2) = 0$) but it has degree zero. – Najib Idrissi May 15 '14 at 10:20
  • Okay, probably I should add some assumption like asphericity. (Which is true for knot complements.) – user39082 May 15 '14 at 11:57
  • I have edited the question accordingly. – user39082 May 15 '14 at 11:58
  • What's the definition of the degree of a map from a manifold with boundary to another? In the case of knot complements, $H_3(S^3\setminus N(K))$ is trivial, so there is no fundamental class. – Dan Rust May 15 '14 at 15:38
  • You consider the relative homology $H_n(M,\partial M)$ which is isomorphic to the integers (for n the Dimension of the manifold). – user39082 May 15 '14 at 17:11
  • For a continuous map $f:M\to N$ the degree is defined via the equation $$f_*\left[M,\partial M\right]=deg(f)\left[N,\partial N\right]$$ – user39082 May 15 '14 at 17:14
  • where $\left[M,\partial M\right]$ is the fundamental class, i.e. a Generator of $H_n(M,\partial M)$. (One needs to fix orientations, otherwise this is only well-defined up to sign.) – user39082 May 15 '14 at 17:15

2 Answers2

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Here's a construction of a degree zero map $\Sigma_2 \to T^2$ such that the induced map on fundamental groups is surjective.

As $\Sigma_2 = T^2\# T^2$, there is a map $f_1 : \Sigma_2 \to T^2\vee T^2$ given by crushing the $S^1$ in the neck.

As $T^2 = S^1\times S^1$, we have a projection map $T^2 \to S^1$, so there is a map $f_2 : T^2\vee T^2 \to S^1\vee S^1$.

Viewing $T^2$ as $[0, 1]^2$ with opposite sides identified, the map $\partial([0, 1]^2) \to [0, 1]^2$ descends to a map $f_3 : S^1\vee S^1 \to T^2$; alternatively, one can view this as the inclusion of the one-skeleton of $T^2$ for the standard CW complex structure on $T^2$.

The composition $f = f_3\circ f_2\circ f_1$ is the desired map $f : \Sigma_2 \to T^2$. It has degree zero because $H_2(S^1\vee S^1; \mathbb{Z}) = 0$. By starting with the standard generators for $\pi_1(\Sigma_2)$ and tracing through the maps, one can see that $f_* : \pi_1(\Sigma_2) \to \pi_1(T^2)$ is surjective. The following image may help to see this.

enter image description here

More generally, provided that $g \geq 2h$, this method can be used to construct degree zero maps $f : \Sigma_g \to \Sigma_h$ such that they are surjective on fundamental groups.

Michael Albanese
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To answer my own question, I have now found the answer in the literature and it is: such maps exist.

Papers by González-Acuna,Ramirez http://www.sciencedirect.com/science/article/pii/S0040938302000873 and Horie,Kitano,Matsumoto,Suzuki http://www.worldscientific.com/doi/abs/10.1142/S0218216511008747 produce examples of epimorphisms between knot groups coming from degree Zero maps.

user39082
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