Compute homology groups of space $\Bbb RP^2$ attached with Mobius band using Mayer Vietories (This is exercise 2.2.28 from Hatcher) Consider the space $X$ (say) obtained from a $\Bbb RP^2$, by attaching a Mobius band $M$ via a homeomorphism from the boundary circle of the Mobius band to the circle $\Bbb RP^1 \subset \Bbb RP^2$. Compute the homology of the space 
We set $A=N_{\epsilon}(\Bbb RP^2)$, $B=N_{\epsilon}(M)$, $X:=A^o\cup B^o $, s.t $A$, $B$ and  $A\cap B$ deformation retracts to $\Bbb RP^2$, $M$ and $ S^1$, so the (reduced) Mayer-Vietoris sequence yields
$$ \tilde H_2(S^1)\to \tilde H_2(\Bbb RP^2)\oplus\tilde H_2(M)\to \tilde H_2(X)\to \tilde H_1(S^1)\to \tilde H_1(\Bbb RP^2)\oplus \tilde H_1(M)\to \tilde H_1(X) \to 0$$
and replacing the homology groups we already know we have
$$0\to 0 \to \tilde H_2(X) \overset{\partial}{\to} \mathbb Z \overset{\phi}{\to} \mathbb Z_2\oplus \mathbb Z\overset{\psi}{\to} \tilde H_1(X)\to 0$$
Now, I have to figure out what the maps $\partial,\phi,\psi$ do, and this is where I'm stuck. I'd appreciate a detailed explanation. 
What I am thinking that $\partial$ is injective and $\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi$. Again $\phi=(j_*,i_*)$ where $i: S^1 \hookrightarrow M , j: S^1 \hookrightarrow \Bbb RP^2$. Then $i_*(1)=2$, but little bit confused about $j_*$ is $j_*(1)=[0]$ because in calculating homology of $\Bbb RP^2$ the map from $e^1 \to e^0$ is zero map again I am thinking that the map might be $1 \mapsto [1]$. Now for $j_*(1)=[0]$.
$\tilde H_1(X)= (\mathbb Z_2\oplus \mathbb Z)/ Im \phi= (\mathbb Z_2\oplus \mathbb Z)/<(2,[0])>=\Bbb Z_2 \oplus \Bbb Z_2$. Then $ker \phi=0=Im \partial$. So, $\tilde H_2(X)=0$. Can anyone please tell me whether I am right or wrong or how to rectify it, moreover what is $ H_0(X)$?
 A: I can see it has been some years since this question was asked, but since it has no answers, let me have a go.
Looking at $0 \mapsto H_{2}(X) \mapsto \mathbb{Z} \mapsto \mathbb{Z}_{2} \oplus \mathbb{Z} \mapsto H_{1}(X) \mapsto 0$.
Let's find $\phi$ first. $\phi$ is induced by the inclusion of $S^{1} \cong \mathbb{R}P^{1}$ into $\mathbb{R}P^{2}$ and its inclusion in $M$. Seeing that $\mathbb{R}P^{1}$ wraps around the mobius band twice we see that $\mathbb{Z} \mapsto\mathbb{Z}/2\oplus \mathbb{Z}$ is $1 \mapsto ([1],2)$. This is injective so $H_{2}=0$. $H_{1}(X)=\frac{\mathbb{Z}/2\oplus\mathbb{Z}}{([1],2)}\cong \mathbb{Z}/4.$ $H_{0}(X)\cong\mathbb{Z}$ being connected.
A: We have that $j$ takes $S^1$ onto $S^1\cong\mathbb{R}P^1\subset\mathbb{R}P^2$, a cycle that represents the generator of $H_1(\mathbb{R}P^2)$, so $j_*(1)=[1]$ and $\text{im}(\phi)=<([1],2)>.$ The kernel of $\phi$ is still zero, so $H_2(X)=0$.
As commented, $H_0(X)=\mathbb{Z}$ since $X$ is path connected.
We will argue $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$. We have the following partition $$\{([k],2k): k\in\mathbb{Z}\}\sqcup\{([k+1],2k): k\in\mathbb{Z}\}\sqcup\{([k],2k+1): k\in\mathbb{Z}\}\sqcup\{([k+1],2k+1): k\in\mathbb{Z}\}\\ = \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}.$$
Under the quotient map $\pi: \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z} \to  (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})/<([1],2)> \cong(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}) / \text{im}(\phi)$,
$$\pi([k],2k)=[([0],0)] $$ $$\pi([k+1],2k)=[([1],0)] $$ $$\pi([k],2k+1)=[([0],1])] $$ $$\pi([k+1],2k+1)=[([1],1)]. $$
With $H_1(X)\cong (\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z})\ /\ \text{im}(\phi)$, the quotient map shows $|H_1(X)|=4$. Since $2[([0],1)]=[([0],2)]$ and $([0],2)=([k+1],2k)$ for $k=1$, we have $2[([0],1)]=[([1],0)]\neq[([0],0)]$ implying $H_1(X)\neq \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}.$ Thus, $H_1(X)\cong \mathbb{Z}/4\mathbb{Z}$, the only other group of order 4.
