$\phi: \pi_1(S)\to \pi_1(S)$ is a homomorphism of fundamental group of closed orientable surface $S$ of genus $\geq 2$. If $\phi$ is not an epimorphism, can we find a non-surjective self map $f: S\to S$ induces $\phi$ such that $f(S)$ be a submanifold embedded in $S$?
2026-03-27 05:03:28.1774587808
The image of homomorphism of fundamental group of closed surface
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I assume that by "submanifold" you mean a submanifold with boundary (otherwise the answer is obviously negative). Then the answer is "yes, this is always possible" to find such $f$.
Proof.
Show that the image $G=\phi(\pi_1(S))$ is an infinite index subgroup in $\pi_1(S)$. Indeed, otherwise it is isomorphic to $\pi_1(S')$, where genus of $S'$ is strictly greater than the one of $S$. However, by looking at the abelianizations, we see that an epimorphism $\pi_1(S)\to \pi_1(S')$ does not exist (otherwise we would obtain an epimorphism $Z^m\to Z^n, m<n$).
It is easy to see that every infinite index subgroup $G$ of $\pi_1(S)$ is free. In our case, this group is also finitely generated. Let $Y$ be the (finite) graph with $\pi_1(Y)=G$. Thus, by Whitehead's theorem, there exists a continuous map $g: S\to Y$ which induces the homomorphism $\phi$. Next, again, by Whitehead's theorem, there exists a PL map $h: Y\to S$ which induces the injection $G\to \pi_1(S)$. Since $g$ is PL, its image is a finite subgraph $Z$ in $S$. Now, let $N$ be a regular neighborhood of $Z$ in $S$: It is a proper subsurface with boundary in $S$. By composing $g\circ h$ we obtain a map $f': S\to N$ which induces the homomorphism $\phi$. The image of $f'$ is just a subgraph, but one can easily perturb $f'$ (in its homotopy class) to a new PL map $f$ so that the image is the entire $N$. This part is not hard but a bit tedious: Let me know if you want to see the details. The point is that there exists a PL map from a 2-simplex onto $N$. Thus, instead of constructing a map $S\to Y$ we first construct a surjective map to $Y$ wedge with the 2-simplex.
Note that the new map $f$ here is far from being injective. It is clear that an injective map would not be possible in general.