Exercise 75.4

Proof of $(a)$. We have the identification of vertices in $P$ as shown in Figure , found by identifying heads/tails of arrows with the same label, where the solid lines are one identification and the dashed lines are another. Thus, $\pi$ does not map all the vertices of $P$ to the same point of $X$. □

Solution for $(b)$. Let $x_0$ be the point identified by the solid lines in Figure , and $x_1$ the point identified by the dashed lines. Then, using Figure , we see that $a$ connects $x_0$ to itself, $c$ connects $x_1$ to itself, and $b$ goes from $x_0$ to $x_1$, while $d$ goes from $x_1$ to $x_0$. Thus, we have the sketch in Figure for $A=\pi (P)$.

We now want to calculate its fundamental group. First, we see that we can homotopically retract the segment $d$ into the point $x_0$, thereby making $x_0$ coincide with $x_1$; the resulting deformation retract is then the wedge sum of three circles. Thus, ${\pi }_1(A,x_0)\simeq {\pi }_1(S^1\vee S^1\vee S^1,x_0)=\mathbb{Z}\ast \mathbb{Z}\ast \mathbb{Z}$ by Theorems $58.3$ and $71.1$. □

Solution for $(c)$. We proceed as in Exercise $\text{(}b)$:

where we cancel $d^{-1}d$ in the penultimate step and relabel in the last step as allowed on p. 460. Thus, our fundamental group is

which has the first homology group

Solution for $(d)$. Since $H_1(X)\simeq {\mathbb{Z}}^2\times (\mathbb{Z}∕2\mathbb{Z})$, by Theorem $75.4$, we have that $X$ is homeomorphic to $P_3=P^2\#P^2\#P^2$, assuming that $X$ is homeomorphic to one of the surfaces in Theorem $75.5$. □