Singular homology of $\mathbb{R}P^{2}$ I am doing algebraic topology course and I'm trying to compute the singular homology for some spaces.
However I am stuck doing that for $\mathbb{R}P^{2}$.
Compute the singular homology of $\mathbb{R}P^{2}$ (the real projective plane).
 A: A perhaps more aesthetically pleasing way is to use that singular and simplicial homology coincides, and then find a triangulation of $\mathbb{R}P^2$.
One such looks like this:  
A: $\mathbb{R}P^2$ is a CW complex with one 0-cell $v$, one 1-cell $a$, one 2-cell $e$ which is attached via the word $a^2$. Cellular homology comes from the chain complex $$0\rightarrow H_2(e,a)\rightarrow H_1(a,v){\rightarrow }H_0(v,\emptyset)\rightarrow 0$$ where $H_i$ are the $i$-th singular homology groups of consecutive pairs of skeletons for the CW structure. Each of these groups in this complex is just a direct sum of copies of $\mathbb{Z}$ with a generator for each $i$-cell. In this case, $$0\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}\rightarrow 0.$$ The boundary maps are computed via the degree of the attaching maps for each cell that generates a $\mathbb{Z}$. In this case the only intersting boundary map is $d_2:[e]\mapsto 2[a]$ where 2 is the degree of the attaching map of $e$ to $a$. All other boundary maps are $0$. So the cellular homology groups (which are the same a singular homology groups) are $H_n(\mathbb{R}P^2)=0$ for $n>1$, $H_1(\mathbb{R}P^2)=\mathbb{Z}_2$, $H_0(\mathbb{R}P^2)=\mathbb{Z}.$  
Cellular homology works very efficiently for computing homology groups of most  familiar CW spaces. i.e., $\mathbb{C}P^n, \mathbb{R}P^n$, genus $g$ surfaces, etc. Hatcher has a lot of examples in the middle of his Homology chapter, and it's very readable.   
A: View the real projective plane as the union of a disk $D^{2}$ and a Möbius strip $M$intersecting in an annulus, so that the edge of the disk goes all the way around the boundary of the Möbius strip. So when we consider $D^{2} \cap M \simeq S^{1}$, the circle "wraps around" the disk twice. Note that $\mathbb{R}P^{2}$ is a connected 2-dimensional manifold so $H_{0}(\mathbb{R}P^{2}) = \mathbb{Z}$ and $H_{n}((\mathbb{R}P^{2}) = 0$ for $n \geq 3$.
Now $M \simeq S^{1}$ and $M \cap D^{2} \simeq S^{1}$, the Mayer Vietrois sequence is
$$H_{2}(D) \oplus H_{2}(S^{1}) \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow H_{1}(S^{1}) \longrightarrow H_{1}(D) \oplus H_{1}(S^{1}) \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow H_{0}(S^{1}) \longrightarrow H_{0}(D) \oplus H_{0}(S^{1}) \longrightarrow H_{0}(\mathbb{R}P^{2})\longrightarrow 0,$$ giving
$$0 \longrightarrow H_{2}(\mathbb{R}P^{2})\longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0.$$
The map including the annulus into the Möbius strip wraps twice around the circle that is a retract of the Möbius strip, so the first map $\mathbb{Z} \rightarrow \mathbb{Z}$ is given by multiplication by $2$. Hence the map is injective and $H_{2}(\mathbb{R}P^{2}) = 0$. Now the last 3 groups in the sequence form a short exact sequence, so the map $H_{1}(\mathbb{R}P^{2}) \rightarrow \mathbb{Z}$ must be the zero map as $\mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}$ is injective. Hence we have a short exact sequence
$$ 0 \longrightarrow \mathbb{Z} \stackrel{2}{\longrightarrow} \mathbb{Z} \longrightarrow H_{1}(\mathbb{R}P^{2}) \longrightarrow 0,$$
so $H_{1}(\mathbb{R}P^{2}) = \mathbb{Z} / 2 \mathbb{Z}$.
