Why can the map between fundamental groups be extended to the whole polygon? 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.
 A: 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!
