# Homology of $\mathbb{R}P^2$ using cellular homology - verification.

As in the title, I want to calculate the homology of $$\mathbb{R}P^2$$ using cellular approach. $$\mathbb{R}P^2$$ has a cell structure with one $$0$$-cell $$x_0$$, one $$1$$-cell $$a$$ and one $$2$$-cell $$e$$, The cellular chain complex is then of the form: $$$$0 \rightarrow \mathbb{Z} \xrightarrow[{}]{d_2}\mathbb{Z} \xrightarrow[{}]{d_1}\mathbb{Z} \rightarrow 0.$$$$ The boundary map $$d_1$$ is zero: we have the composition $$$$S^0 \xrightarrow[{}]{\gamma} \{x_0\} \rightarrow \{x_0\}/\emptyset \xrightarrow[{}]{=} S^{0},$$$$ and the degree of this composition is $$0$$. Then $$d_1$$ is trivial, as claimed. For $$d_2$$, we consider: $$$$S^1 \xrightarrow[{}]{\gamma} S^1/\sim=\mathbb{RP^1} \xrightarrow[{}]{q} \mathbb{RP^1}/\mathbb{RP^0} \cong S^{1}.$$$$ The attaching map $$\gamma$$ is given by $$a^2$$ and so the composition above has degree $$2$$. We then end up with: $$$$0 \rightarrow \mathbb{Z} \xrightarrow[{}]{2}\mathbb{Z} \xrightarrow[{}]{0}\mathbb{Z} \rightarrow 0.$$$$ Now, $$H_2 \cong 0$$ since $$\times 2$$ map is injective in $$\mathbb{Z}$$ and so its kernel is zero. Putting all the above together, we have that $$H_0 \cong \mathbb{Z}, H_1 \cong \mathbb{Z}/2\mathbb{Z}$$ and $$H_i \cong 0$$ for $$i > 1$$. Is that a correct argument?

• Looks good to me. You might want to argue why, at least, $H_2(\mathbb{R}P^2) \cong 0$ follows from your cellular chain complex, but this is quick since multiplication by $2$ is injective in $\mathbb{Z}$. Commented Dec 18, 2019 at 13:40
• @kamills thanks! I have edited the answer as per your suggestion. Is that what you had in mind? Commented Dec 18, 2019 at 14:06
• Not quite--this sequence is not exact (otherwise it would have zero homology everywhere). The homology of $\mathbb{R}P^2$ is given by the homology of your cellular chain complex. My point was that, to calculate $H_2$, you have to compute the kernel of multiplication by 2, but multiplication by 2 is injective so the kernel is zero, hence $H_2 \cong 0$. Commented Dec 18, 2019 at 14:08
• @kamills oh ok, I see. I got confused there. Thanks again for your help! Commented Dec 18, 2019 at 14:11
• @kamills If you wanted to turn your comments into an answer, I’d happily (+1) Commented Dec 18, 2019 at 14:57

Everything is good up through the cellular chain complex, given below, which is only nonzero in degrees zero through two:

$$\cdots \to 0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$$

Then you had the right answer, and a comment I made caused a bit of confusion which led me to point out something important: the cellular chain complex is (in general) not exact, and computing its homology gives you the homology of your space. So everything you added in your edit starting with "The exact sequence of homology groups is..." is not how you'll want to finish it off (my comment had meant something more like, "finish it off by adding a sentence for how you can tell $$H_2 \cong 0$$ when you're reading off the homology of your chain complex"). Edit: The OP got edited so this above paragraph is more or less irrelevant.

Anyways, now let's compute the homology of $$\mathbb{R}P^2$$, which amounts to the homology of the above chain complex. Label the maps $$\delta_1$$ and $$\delta_2$$. Then $$H_0(\mathbb{R}P^2) \cong \mathbb{Z}/\text{im }\delta_1 \cong \mathbb{Z}/0 \cong \mathbb{Z}$$.

$$H_1(\mathbb{R}P^2) \cong \ker\delta_1/\text{im }\delta_2$$. Since $$\delta_1$$ is zero its kernel is all of $$\mathbb{Z}$$, and since $$\delta_2$$ is multiplication by $$2$$ its image is $$2\mathbb{Z}$$, so $$H_1(\mathbb{R}P^2) \cong \mathbb{Z}/2\mathbb{Z}$$.

$$H_2(\mathbb{R}P^2) \cong \ker\delta_2/\text{im }0 \cong \ker \delta_2$$. Since $$\delta_2$$ is multiplication by $$2$$, it is injective ($$\mathbb{Z}$$ is an integral domain), so $$\ker \delta_2 = 0$$. Thus $$H_2(\mathbb{R}P^2) \cong 0$$.

Since the cellular chain complex consists of zeros above degree $$2$$, $$H_i(\mathbb{R}P^2) \cong 0$$ for $$i > 2$$ as well.