# Computing map on Cohomology induced by inclusion.

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.

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.