Exercise 1.2.9

Proof. Suppose $C$ is a retract. Proposition $1.17$ implies ${\iota }_{\text{∗}}$ is injective. We claim ${\tilde{\iota }}_{\text{∗}}$ is also. Since ${\iota }_{\text{∗}}$ is injective, ${\tilde{\iota }}_{\text{∗}}(x)=0$ for $x\in {\pi }_1(C)$ if and only if $x$ is in the derived subgroup of ${\pi }_1(M_h^{\prime })$. But this implies that $x=0$, for any element in the derived subgroup of ${\pi }_1(M_h^{\prime })$ is of the form $\mathit{ab}{a}^{-1}{b}^{-1}$, and the only element in the image of ${\iota }_{\text{∗}}$ of this form is ${\iota }_{\text{∗}}(0)=1$, since ${\pi }_1(C)=\mathbb{Z}$. Thus, ${\tilde{\iota }}_{\text{∗}}={\iota }_{\text{∗}}\circ q$ is injective.

We claim this is a contradiction. We first draw the cell structure of $M_h^{\prime }$:

Note that $c$ is the generator of ${\pi }_1(C)$. We see that the 2-cell is attached using the scheme $c[a_1,b_1]\cdots [a_g,b_g]$, and so the fundamental group is given by

which has the abelianization

since ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^g[a_i,b_i]\mapsto 0$ under the abelianization map. This implies that ${\tilde{\iota }}_{\text{∗}}(c)=0$, contradicting injectivity. Thus, $M_g$ does not retract onto $C$, for the restriction of this retraction to $M_h^{\prime }$ would then be a retract $M_h^{\prime }\to C$.

We now claim that $M_g$ retracts onto the nonseparating circle $C^{\prime }$. We consider part of our cell structure in the following diagram:

where $a_i$ is identified with the circle $C^{\prime }$. We claim that we can extend a retraction projecting the top copy of $a_i$ to the bottom copy in the square on the right side can be extended to a retraction of the whole complex. This retracts the left dashed edge of the square into one point $p$, and the same applies to $b_i$ on the other side of this square. We can map the rest of the polygon outside of the right square onto $p$; this is a well-defined map since every edge not equal to $a_i$ maps to $p$ and the edge $a_i$ stays constant throughout our map. This is moreover a continuous map since an open subset $U$ of $a_i$ has preimage of a union of open vertical slices of the right square if it does not contain $p$, and if it does, has preimage another union of open vertical slices unioned with the rest of the polygon on the left. □