Why can the map between fundamental groups be extended to the whole polygon?

Question

Let $M$ be a surface with genus $g$. Suppose there is an injection $\pi_1(M) \to \pi_1(N)$, where $N^2$ is compact and oriented. We can view $M$ as a $4g$-sided polygon with appropriate identification and defining a boundary map via the inclusion $\pi_1(M) \to \pi_1(N)$. I am wondering why this allows me to extend the map $\phi:M \to N$. The hint I am given is to use contractibility, but I am not very sure how to use this hint.

Answer 1

Let's call

• $P$ your $4g$-gon;
• $f:P → M$ the quotient map;
• $\hat\gamma:[0,1] → \partial P$, a simple loop circumnavigating $\partial P$;
• $\gamma= f\circ \hat\gamma$, the image onto $M^1$; and
• $a_1,b_1,\ldots,a_g,b_g:S^1→ M^1$ the loops generating $\pi_1(M)$, such that $$\gamma = [a_1,b_1][a_2,b_2]\cdots[a_g,b_g].$$

Let $\phi_\ast$ denote your homomorphism, and $\phi$ the map to be derived.

Choosing representatives of the images of the generators $\phi_\ast(a_i),\ \phi_\ast (b_i)$ gives a map $M^1→N$ as you mention: $$\phi:\bigcup a_i\cup b_i → \bigcup \phi(a_i)\cup \phi(b_i) \tag(1)$$ The fact that $\phi_\ast$ is a homomorphism then tells us that

$$[\phi(a_1),\phi(b_1)][\phi(a_2),\phi(b_2)]\cdots[\phi(a_g),\phi(a_g)]$$

is nullhomotopic in $N$, since it represents the class

$$\phi_\ast([a_1,b_1][a_2,b_2]\cdots[a_g,b_g]) = \phi_\ast (1_{\pi_1(M)}) = 1_{\pi_1(N)}.$$

Let $F: S^1 × [0,1] → N$ denote the nullhomotopy. I.e.:

• $F$ is continuous.
• $F(\cdot, 1) = [\phi(a_1),\phi(b_1)][\phi(a_2),\phi(b_2)]\cdots[\phi(a_g),\phi(a_g)]$.
• $F(\cdot, 0)$ is constant.

By the last bullet point, $F$ passes to the quotient to give a map

$$F: \frac{S^1 × [0,1]}{S^1 × \{0\}} → N,$$

Theis new domain of $F$ is, up to homeomorphism your polygon (a.k.a. disk): i.e. there exists a homeomorphism

$$\Psi:P → \frac{S^1 × [0,1]}{S^1 × \{0\}}.$$

One just has to show that $\Psi$ can be chosen so that $F\circ\Psi: P \to N$ passes again to the quotient $f$ to give a continuous map $M → N$.

This is a condition only on the restriction of $\Psi$ from boundary to boundary: $\partial P → S^1 × \{1\}$. Since $\gamma$ is simple, we can take

$$\Psi^{-1}(x,1) = \gamma(x)$$

and this does the trick nicely—from the second point about $F$, it follows that $\Psi \circ F$ restricted to $\partial P$ is equal to the function $\phi$ as in $(1)$, and so easily passes to the quotient, as required!