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I am not very comfortable with computations related to cohomology and I am having trouble with part (c) of this exercise. I successfully computed the Cohomology of $D$ with $\mathbb{Z}$ and $\mathbb{Z}_2$ coefficients using Mayer Vietoris for homology and UCT. $H_1(D) = \mathbb{Z}_2 \oplus \mathbb{Z}_2$, $H_2(D) = 0$, $H_0(D) = \mathbb{Z}$. I give the results I obtained using UCT here in case I made a mistake. $H^2(D) = \mathbb{Z}_2 \oplus \mathbb{Z}_2$ and $H^1(D) = \mathbb{Z}^2$ and $H^(0)(D) = \mathbb{Z}$. Now I want to compute the map induced by the inclusion of $\mathbb{R}P^2 \rightarrow D$ with integer coefficients on cohomology. I was not really sure how to start doing that, as I cannot think of what a generator gets mapped to. I started thinking about understanding the map induced on homology instead, focusing on $H_1(\mathbb{R}P^2)=\mathbb{Z}_2 \rightarrow H_1(D)$, I could not really make much progress here either, so I thought that maybe relative homology might help. I.e I was thinking about considering the space obtained when $D$ is quotiented by $\mathbb{R}P^2 \subset D$, by identifying it to a point. The resulting space would be $S^2$, so the sequence of relative homology gives $$... H_1(\mathbb{R}P^2)=\mathbb{Z}_2\rightarrow H_1(D) = \mathbb{Z_2}\oplus Z_2 \rightarrow H_1(S^2) = 0 ...$$ However this seems to imply that the map induced by the inclusion has image $\mathbb{Z}_2\oplus \mathbb{Z}_2$ by exactness. Something must be wrong here...

But besides that my main question, is basically how to do part $c)$ of the exercise. I would appreciate a full calculation, as I am not comfortable with the machinery involved.

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The mistake is in saying that the space formed when the the boundary of $M$ and $\mathbb{R}P^2$ are collapsed to a point in $D$ gives $S^2\times S^2$. It actually gives $\mathbb{R}P^2\times S^2$. Hence in my sequence I have $\mathbb{Z}_2\xrightarrow{f}\mathbb{Z}_2\oplus \mathbb{Z}_2 \xrightarrow{\phi} \mathbb{Z}_2$. As I argued before $\phi$ cannot map to $0$, so $\phi$ maps onto $\mathbb{Z_2}$. The only possibility is that one of $(0,1)$ and $(1,0)$ gets sent to $1$ and the other to $0$, ie a projection. This that kernel $(0,0)$ and $(0,1)$ or $(1,0)$, ie $\mathbb{Z}_2$, hence $f$ has image $\mathbb{Z}_2$. So $f(1) = (1,0)$ is all we have. When considering the map induced on $H_(\mathbb{R}P^2) \rightarrow H_2(D)=0$, the map has to be trivial. The map on cohomologies then follow.

I am still thinking about the computation of the cup product.

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