Homology of $S^2$ with relation $x ∼ -x$ (First part Hatcher 2.2 Exercise 10)

algebraic-topologyhomology-cohomology

I'm trying to solve Exercise 2.2.10 from Hatcher. I already solved this Exercise once without any trouble, but now that I'm preparing for my test, for the life of me, I can't figure out what I'm doing wrong. Specially $H_1(X)$ is not correct. I can solve this with a CW-complex no problem. If somebody can tell me what mistake I'm making I would be extremely thankful!

Let $X$ be the quotient space of $S^2$ under the identifications $x ∼ −x$ for $x$ in the equator $S^1$.
Compute the homology groups $H_i(X)$.

I want to solve this using Mayer Vietoris. Take $A, B$ two open sets which deform and retract onto the northern and southern hemispheres. These fulfill conditions for Mayer Vietoris and $A\cup B = X$, $A\cap B \cong RP^1$ and $A, B \cong RP^2$. Using $H_i(RP^1)= \mathbb{Z},\mathbb{Z},0$ for $i=0,1$ and rest, and $H_i(RP^2)= \mathbb{Z}, \mathbb{Z}_2, 0$ for $i=0,1$ and rest, we get the following relevant part of the MV sequence, utilizing that $RP^1$ is path connected:

$ 0 \rightarrow H_2(X) \rightarrow H_1(RP^1) \xrightarrow{\varphi_*} H_1(RP^2)\bigoplus H_1(RP^2) \xrightarrow{\psi_*} H_1(X)\rightarrow 0$.

$ 0 \rightarrow H_2(X) \rightarrow \mathbb{Z} \xrightarrow{\varphi_*} \mathbb{Z_2}\bigoplus \mathbb{Z_2}\xrightarrow{\psi_*} H_1(X)\rightarrow 0$.

Note that $\varphi_*$ is a map from the free abelian group $\mathbb{Z}$ to $\mathbb{Z_2}\bigoplus\mathbb{Z_2}$, so has to be trivial. It follows $H_2(X)\cong \mathbb{Z}$.
Note as well that $\psi_*$ is surjective so it holds $H_1(X)\cong \frac{\mathbb{Z_2}\bigoplus\mathbb{Z_2}}{Im \psi_*}\cong \frac{\mathbb{Z_2}\bigoplus\mathbb{Z_2}}{ker \varphi_*}\cong \mathbb{Z_2}\bigoplus\mathbb{Z_2} $.

But $H_1(X)$ is supposed to be only one copy of $\mathbb{Z_2}$!

What am I doing wrong?

Thanks in advance

Best Answer

If $i:A\cap B\to A$ and $j:A\cap B\to B$ are inclusions, then $i_*(1)=1$ and $j_*(1)=1$ on the first homology groups. By definition of $\varphi_*$ given in the Wikipedia link that I gave in the comments, we have $$\varphi_*(1)=(i_*,j_*)(1)=(1,1)$$ in the exact squence that you considered above and hence

$H_1(X)\cong \frac{\mathbb{Z_2}\bigoplus\mathbb{Z_2}}{Ker \psi_*}\cong \frac{\mathbb{Z_2}\bigoplus\mathbb{Z_2}}{Im \varphi_*}\cong \mathbb{Z_2}$.