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.
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.
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.