Cohomology of space obtained by identifying the boundary of $M$=mobius band to $\mathbb{R}P^1\subset \mathbb{R}P^2$ using Mayer Vietoris. 
I am struggling with a particular part of this question. I think that I can do a) b)i) b)iii) ( provided I have b)ii) and I use UCT). My problem is with the computation of b)ii): I can compute $H^0(D)$ and $H^(1)(D)$ but i can't seem to compute $H^2(D)$ because in the long exact sequence  ( Mayer vietoris for cohomology), $H^2(D)$ is sandwiched between $H^1(M\cap \mathbb{R}P^2) = H^1(S^1)= \mathbb{Z}$ and $H^2(\mathbb{R}P^2)\oplus H^2(S^1) = \mathbb{Z}_2$ and all I managed to figure out is the corresponding map from $H^2(D)$ to $H^2(\mathbb{R}P^2)\oplus H^2(S^1)$ is surjective. I give more details below:
I used Mayer Vietoris long exact sequence with $X$ being the union of the interiors or $A$ and $B$:
$$H^n(X)\rightarrow H^n(A)\oplus H^n(B) \rightarrow H^n(A\cap B) \rightarrow H^{n+1}(X)...$$
Here $X=D$ and I chose $A$ to be $M$ and $B$ to be the union of $\mathbb{R}P^2$ with a small chunk on $M$ that deformation retracts to $\mathbb{R}P^2$. Here I give the section to compute $H^2(D)$:
$$H^1(M\cap \mathbb{R}P^2)=H^1(S^1) \xrightarrow{\partial} H^{2}(D)\xrightarrow{\phi} H^2(M)\oplus H^2(\mathbb{R}P^2)=\mathbb{Z}_2 \rightarrow H^2(M\cap \mathbb{R}P^2)=0$$
This tells me, by exactness that $\phi$ is surjective. In order to compute $H^2(D)$, I would need $ker(\phi) = Im(\partial)$. But i can't seem to figure out how to compute this part. I would appreciate a a solution for this calculation. 
I have seen solutions for the calculation for the homology and I do understand those but I am still stuck here. 
 A: You haven't actually stated what your CW or simplicial structures are. Let's try doing everything with CW structures. They're slightly more flexible and they subsume simplicial homology. 
The CW structure on $\mathbb{R}P^2$ is pretty straightforward. It's $e^0 \cup 
e^1 \cup e^2$, where the $e^2$ is glued on by a degree $2$ map. Hatcher has an explanation for this. Note that the "standard circle" $\mathbb{RP}^1$ is represented exactly by the $e^1$ in this decomposition. 
The cellular chain complex is $0 \to \mathbb{Z} \xrightarrow{\times 2} \mathbb{Z} \xrightarrow{\times 0} \mathbb{Z} \to 0$. If you're not sure how these maps arise, look at Hatcher. So we get $H_2(\mathbb{R}P^2) = 0, H_1(\mathbb{R}P^2) = \mathbb{Z}/2, H_0(\mathbb{R}P^2) = \mathbb{Z}$. Using UCT, this gives $H^2(\mathbb{R}P^2) = \mathbb{Z}/2, H^1(\mathbb{R}P^2) = 0, H^0(\mathbb{R}P^2) = \mathbb{Z}$.
Great, now let's find a CW structure on $M$. If you have the fundamental polygon, you can see that you have $2e^0s, 3e^1s, 1e^2s$. I'm labelling mine as below. I'm going to use the convention (which is not necessary for CW but is for simplicial) that my edges/$e^1$s are oriented from $v_1$ to $v_2$. So, the fundamental polygon is equivalent to the CW complex with a $2$ cell attached along the word $cb^{-1}ca^{-1}$. Note that the boundary of $M$ is $a-b$. 


So, let's try to get a cell structure for $D$ from this. We can map the $e^0$ of $\mathbb{R}P^2$ to $v_1$, and then since we are identifying the boundary circle with $e^1$, we have $e^1$ mapping to $ab^{-1}$ and so the $e^2$ wraps around $ab^{-1}$ twice, meaning, that it gets attached via the map $ab^{-1}ab^{-1}$. 
So this is our cell structure on $D$: $2 e^0s$, $3e^1s$ as before, and $2e^2$, attached by $cb^{-1}ca^{-1}$ and $ab^{-1}ab^{-1}$. Since we are dealing with (co)homology, only the abelianized relations will matter, so we can say we are gluing by $a + b - 2c$ and $2a + 2b$. 
Great, now our cellular chain complex for $D$ is
$$0 \to \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{\partial_2} \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \xrightarrow{\partial_1} \mathbb{Z} \oplus \mathbb{Z} \to 0$$
Let us calculate each homology group.
For $H_2(D)$, we note that $\partial_2$ maps $(1,0)$ to $(1,1,-2)$ and $(0,1)$ to $(2,2,0)$. This map is injective, so the kernel is empty so $H_2(D) \cong 0$.
For $H_1(D)$, we note that $\partial_1$ maps $a = (1,0,0),b = (0,1,0),c =(0,0,1)$ all to $(1,-1)$. So the kernel is generated by $a-c, b-c$. Notice that the image of $\partial_2$ says that $a-c+b-c = 0$. It also says that $2a-2b = 0$. So We get $$H_1(D) \cong \mathbb{Z}\{a-c, b-c\}/(a-c+b-c, 2a-2b) \cong \mathbb{Z}\{a-c\}/(a-c+a-c) = \mathbb{Z}\{a-c\}/(2(a-c)) = \mathbb{Z}/2$$
Last, for $H_0(D)$, we note that the boundary sends everything to $0$, so the kernel is everything. The image of $\partial_1$ says $v_2 = -v_1$, so we get $H_0(D) \cong \mathbb{Z}\{v_1,v_2\}/(v_1 = -v_2) = \mathbb{Z}$. 
This avoids using Mayer-Vietoris entirely. Now you can use UCT to get cohomology with $\mathbb{Z}$ coefficients, and cohomology with $\mathbb{Z}/2$ coefficients.
A: Let us try Mayer-Vietoris in homology and look at $H_1$ (I think this is essentially the same place you are stuck). You have
$H_2(D) \xrightarrow{\partial_{2*}} H_1(S^1) \xrightarrow{s} H_1(\mathbb{R}P^2) \oplus H_1(M) \to H_1(D) \xrightarrow{\partial_{1*}} H_0(S^1) \to H_0(\mathbb{R}P^2) \oplus H_0 (M) \to H_0(D) \to 0$ 
I think you can argue that $\partial_{1*}$ is actually $0$, since the map $H_0(S^1) \to H_0(\mathbb{R}P^2) \oplus H_0(M)$ is injective. Great so we actually have a sequence $H_1(S^1) = \mathbb{Z} \to \mathbb{Z}/2 \oplus \mathbb{Z} \to H_1(D) \to 0$. For similar reasons the map $\partial_{2*}$ is $0$ too, because $H_1(S^1) \to H_1(\mathbb{R}P^2) \oplus H_1(M)$ is injective (it maps a loop in $S^1$ to the exact same loop in the core circle in $H_1(M)$). So I have a short exact sequence $0 \to \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}/2 \to H_1(D) \to 0$. This sequence is actually split, because a loop in $S^1$ gets mapped to the "same" loop in $M$, so in particular we have a map $$t:H_1(\mathbb{R}P^2) \oplus H_1(M) \to H_1(S^1)$$ such that $s \circ t$ is the identity on $H_1(S^1)$. This means the sequence is left split, so indeed we have $H_1(D) = \mathbb{Z}/2$. In fact this calculation made me realize I had an error in my calculation with cellular homology that I am just about to fix. 
Calculating $H_2$ using MV should be trivial (since $H_2(S^1) = H_2(\mathbb{R}P^2) = H_2(M) = 0)$, and so we get the same homology results. Now you can UCT and continue. 
