may be a typo in the solution of problem 10, part(a) in section 2.2 of AT. The problem and its solution are given below:




My questions are:
1- Is not this space is the real projective plane? if so, then the homology groups of the real projective plane are calculated on pg.106 in AT by simplicial homology and in that case $H_{2} = 0,$ so why in the solution it is $\mathbb{Z}$? if this is not the real projective plane, then why $H_{2} = \mathbb{Z}$
2-Also, what is the formula used in calculating $\partial_{2}$ is it the cellular boundary formula? if so where is the space $X^{(n-1)}$?
EDIT:
1-could someone explain the solution for me please in an answer?
 A: Question 1: No, this space is not $\mathbb{R}P^2$. The answer given by William here describes the space.
Question 2: Hatcher gives the cellular boundary formula in Section 2.2. I'm going to try to explain it with less heavy notation, but this means I'll have to use more words. The cellular boundary $\partial_n: C_n^{cell} \to C_{n-1}^{cell}$ in the cellular chain complex for a space $X$ is given by the following construction. Remember that these cellular chain groups are free abelian with basis the $n$- (resp. $(n-1)$-)cells of $X$. So $\partial_n$ is a matrix. To determine a formula for $\partial_n$ we need to figure out what $\partial_n$ is on each $n$-cell, and it will be given by some $\mathbb{Z}$-linear combination of the $(n-1)$-cells. Here's how we do it.
First, the attaching map for an $n$-cell is a map $S^{n-1} \to X^{n-1}$ where $X^{n-1}$ is the $(n-1)$-skeleton of $X$. Then quotienting out the $(n-2)$-skeleton we get a map $X^{n-1} \twoheadrightarrow X^{n-1}/X^{n-2}$, but this image is homeomorphic to $\bigvee S^{n-1}$ , a wedge of spheres (one for each $(n-1)$-cell). Now we can collapse all these spheres, except for one, and that one will correspond to one of the $(n-1)$-cells. The degree of this composition is the coefficient in front of that cell in the formula for $\partial_n$.
So, to recap: $\partial_n$ is given by the matrix of numbers corresponding to the degrees of
$S^{n-1} \xrightarrow{\text{attach $n$-cell}} X^{n-1} \twoheadrightarrow X^{n-1}/X^{n-2} \cong \bigvee S^{n-1} \xrightarrow{\text{collapse all but single $(n-1)$-sphere}} S^{n-1}$
for the various $n$-cells and $(n-1)$-cells. If you want the full formula, with the indexed notation, read it in Hatcher. The point is that this formula really boils down to looking at the attaching maps and seeing what they do when restricted to each cell.
I'll leave you to figure out how this corresponds to $\partial_2$ for your space, since it reduces to replacing $n$ by 2 in everything I've written above and then looking at the description of the space in the linked answer plus the answer in your picture.
Question 3: You asked how this could be done by Mayer-Vietoris. Here's a hint:
We can decompose $X$ into sets $U = D_+ \cup N$ and $V = D_- \cup N$ where $N$ is an open collar around the circle along which $D_+$ and $D_-$ get glued. Then $U \simeq D_+$, $V \simeq D_-$, and $U \cap V \simeq S^1$. But then $D_+$ and $D_-$ are contractible. What does this give you for your Mayer-Vietoris exact sequence?
