# Compute homology groups

I'm not confident with this kind of problem, so I post my solution here to ask for checking it. I want to compute the homology groups of the space obtained from two copies of $$\mathbb{R} P^2$$ by gluing them along standard copies of $$\mathbb{R} P^1$$.

First I give it a cell structure: we know that $$\mathbb{R}P^2$$ has the cell structure of one $$0$$-cell $$x$$, one $$1$$-cell $$a$$ and one $$2$$-cell $$A$$ that glues to $$2a$$ (i.e. go around $$a$$ 2 times). So I believe the cell struture of our space is one $$0$$-cell $$x$$, one $$1$$-cell $$a$$ and two $$2$$-cells $$A,B$$ that both glue to $$2a$$. Hence we have the chain complex $$0\to \mathbb{Z}^2 \xrightarrow{d_2} \mathbb{Z} \xrightarrow{d_1}\mathbb{Z} \to 0$$ where $$d_1=0$$ and $$d_2(A)=d_2(B)=2a$$. Hence $$H_0=\mathbb{Z}/0=\mathbb{Z}$$ $$H_1=\langle a\rangle /2\langle a \rangle = \mathbb{Z}_2$$ $$H_2=\langle A-B\rangle / 0 = \mathbb{Z}$$

Is this correct?

• Seems fine. Just be careful with your terminology: you wrote down the chain complex, not an exact sequence. The homology groups of an exact sequence are all zero! – Ethan Dlugie May 3 at 6:08
• edited. Thank you. I am really worried about the cell structure. One move wrong and everything goes wrong. – Marcos G Neil May 3 at 6:13

Let $$A, B, C$$ be CW complexes, and let $$f\colon A \to B$$ and $$g\colon A \to C$$ be cellular maps. Then if $$f'$$ is homotopic to $$f$$ there is a homotopy equivalence $$B\cup_{f, g} C\simeq B\cup_{f',g}C$$.
Now define $$D$$ as the quotient $$\mathbb{RP}^2 \cup_{\iota, \iota} \mathbb{RP}^2$$ where $$\iota\colon \mathbb{RP}^1\to \mathbb{RP}^2$$ is the standard inclusion. This has a cell structure as you've described. We can think about it as starting with a copy of $$\mathbb{RP}^2$$ and attaching a $$2$$-cell via a map $$\varphi\colon\partial D^2 \cong S^1\to \mathbb{RP}^1\cong S^1$$ of degree $$2$$, i.e. $$D$$ is homeomorphic to $$\mathbb{RP}^2\cup_{\varphi} D^2$$.
However, since $$\pi_1(\mathbb{RP}^2)\cong \mathbb{Z}/2$$ this attaching map is null-homotopic so in fact $$D$$ is homotopy-equivalent to $$\mathbb{RP^2}\vee S^2$$ by the above lemma (cf this related question), whose homology groups are easily computable via $$\tilde{H}_k(A\vee B)\cong \tilde{H}_k(A) \oplus \tilde{H}_k(B)$$.
• Yes, I'm writing reduced homology at the end because the additivity statement isn't true in degree $0$. If $r_1=rank (H_0(A))$ and $r_2=rank (H_0(B))$ then $rank (H_0(A\vee B) )= r_1 + r_2 -1$ (since you're wedging together the two components containing the base points). – William May 3 at 16:35