Monoid of continuous self-maps of (real) surfaces

Question

Let $S$ be a closed surface of genus $g > 0$ and $[S,S] = Hom(\pi_{1}(S),\pi_{1}(S))$ be the monoid of (homotopy classes of) continuous maps from $S$ to itself. Consider the semi-group $A$ of elements that induce the $0$ map on $H_{2}(S,\mathbb{Z})$.

Question. Is there a "nice" set of generators for $A$? (Maybe we quotient by natural conjugacy action of the homeomorphisms of $S$.)

The map on $\mathbb{S}^{1} \times \mathbb{S}^{1}$, $(x,y) \rightarrow (x,1)$ would be an example of such a map on the torus, I guess that this generates up to conjugacy?

EDIT

More generally, we can generate many classes in $A$ by picking any continuous mapping $f: S \rightarrow \mathbb{S}^1$ (of which there are many different classes; precisely $\mathrm{Hom}(\pi_{1}(S),\mathbb{Z})$), and choosing some continuous mapping $\mathbb{S}^{1} \hookrightarrow S$. More specifically, do maps of the above form generate $A$?

(This seems quite unlikely at the algebraic level, but if there is a counterexample I would be interested to see a geometric realisation of the map as well)

Answer 1

Here's some sort of geometric description of maps of degree 0. I don't know how realistically you can find a generators-and-relations description of the semigroup of such maps.

It is a theorem due to Kneser that I learned from this answer that any map of degree 0 between closed oriented surfaces is homotopic to a non-surjective map. So $\Sigma_g \to \Sigma_h$ factors through the 1-skeleton up to homotopy. You're reduced to asking for a classification of maps $\Sigma_g \to \vee_1^{2h} S^1$ up to homotopy (which again is equivalent to the algebraic statement about maps on the fundamental group). The largest free group $\pi_1(\Sigma_g)$ surjects onto is $F_g$. To see this, look at the induced map on cohomology of the corresponding Eilenberg MacLane spaces; the fact that the map is a surjection implies that $\Bbb Z^k \cong H^1(F_k) \to H^1(\Sigma_g)$ is injective. The cup product must be zero on the image; because it's a nondegenerate bilinear form on $H^1(\Sigma_g)\cong \Bbb Z^{2g}$, the image can be at most $g$-dimensional.

Dualizing, we see that when we have a surjection $\pi_1(\Sigma_g) \to F_g$, the kernel of the map on $H_1$ is $\Bbb Z^g$, which furthermore can be given a basis of pairwise disjoint simple closed curves with connected complement.

I'd like to be able to say that this is true further at the level of fundamental groups, so that (after precomposing with a diffeomorphism of the domain) every map $\Sigma_g \to \Sigma_g$ of degree 0 extends over the handlebody $H_g$. (Note that algebraically this semigroup is not at all generated by the maps that factor through $S^1$ - the image of the fundamental group of $gf$, where $f$ is such a map, necessarily has image a quotient of $\Bbb Z$.) But I don't know how to get the statement at the level of fundamental groups, though I think it's true.

Answer 2

This semigroup does not have to be finitely generated (if you don't allow up to conjugacy). As @YCor points out for a torus we are looking at $2\times 2$ integer matrices of determinant 0. If $X$ is a finite set of such matrices, then the image of any element of the subsemigroup generated by $X$ as an operator on $\mathbb Z\times \mathbb Z$ must be contained in the image of an element of $X$. Since the non-zero matrices of $X$ have $\mathbb Q$-rank 1 if $X$ generated the semigroup it would imply that $\mathbb Q ^2$ is a union of finitely many lines (the ones spanned by the images of the nonzero elements of $X$), a contradiction.