TITLE: Surface of genus $g$ does not retract to circle (Hatcher exercise)
QUESTION [11 upvotes]: I'm trying exercise 9 on page 53 in Hatcher but I need some help with it. The exercise is:
In the surface $M_g$ of genus $g$, let $C$ be a circle that separates $M_g$ into two compact subsurfaces $M_h^\prime $ and $M_k^\prime$ obtained from the closed surfaces $M_h$ and $M_k$ by deleting an open disk from each. Show that $M_h^\prime$ does not retract onto its boundary circle $C$, and hence $M_g$ does not retract onto $C$. [Hint: abelianize $\pi_1$.] But show that $M_g$ does retract onto the nonseparating circle $C^\prime$ in the figure.

My first question is: assume there wasn't the hint, how would I think of abelianising? What does it mean exactly?
I thought I could do this by contradiction: if it retracts the induced map $i_\ast : \pi_1(C) \rightarrow \pi_1(M_h^\prime) $ is injective. I know $\pi_1(C) \cong \mathbb{Z}$, then I computed $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and then I think I'm stuck. Right? Do you agree with $\pi_1(M_h^\prime) \cong \mathbb{Z}$ and being stuck after that?
What do I need to know to make progress? Many thanks for your help!

REPLY [4 votes]: Here is an approach using homology to show that $Mg$ does not retract to $C$:
Use the fact (proven here: Homology of surface of genus $g$), that $H_1(Mg, \mathbb Z) =\mathbb Z^{2g}$, and that $H_1(C,\mathbb Z)=\mathbb Z$ . Then we use the standard method of composing the retraction with the inclusion:
If there was a retraction $r: Mg\rightarrow C $, then the composition $roi$ with the inclusion $i$ would give an isomorphism h:$\mathbb Z \rightarrow \mathbb Z$, which cannot happen, because of the intermediate group $\mathbb Z^{2g}$.