S.E. Mathematicians,
I have trouble solving the following problem:
Let $M = \mathbb{C}P^3$ with a point removed. Determine the cohomology group $H^i (M,\mathbb{Z})$ and compact support cohomology $H^i _c (M,\mathbb{Z})$ for $i = 0,1,2,3$. Also, determine the natural map from $H^i _c (M,\mathbb{Z}) —> H^i (M,\mathbb{Z})$ for $i = 0,1,2,3$. The $\mathbb{Z}$ is the set of all integers.
My line of attack is using the Mayer-Vietoris sequence for computation: Express $\mathbb{C}P^3$ as a union of $M$ and a 6-Ball $B$, where the special point $c$ is deleted from $\mathbb{C}P^3$ resides in the interior of $B$. The homology of $B$ and $B\setminus\{c\}$ is mostly trivial, so the homology of $M$ usually equals that of $\mathbb{C}P^3$. That means $H^i (M,\mathbb{Z})$ is isomorphic to $H^i (B,B\setminus\{x\},\mathbb{Z})$, and $H^i _c (M,\mathbb{Z})$ is isomorphic to trivial group...right? I do not even know how to determine the natural map....
$\mathbb{C}P^3$ can be constructed as attaching a 6-cell via the quotient map $S^5 \rightarrow \mathbb{C}P^2$. Hence, if we remove the tip of the cone, we can deformation retract the resulting space onto $\mathbb{C}P^2$. So the cohomology is that of $\mathbb{C}P^2$ (this is not the same as what you claim). Compactly supported cohomology is the same as the reduced cohomology of the one point compactification. The one point compactification of a compact manifold minus a point is the compact manifold you started with. Hence the compactly supported cohomology is the reduced cohomology of $\mathbb{C}P^3$.
The map they are talking about from compactly supported cohomology to cohomology is the inclusion of the compactly supported cochains into all cochains. Ignoring dimension 0 issues, this is the same as the map on unreduced cohomology induced by including $\mathbb{C}P^3 - *$ into $\mathbb{C}P^3$. The argument above shows that this is equivalent to including $\mathbb{C}P^2 \rightarrow \mathbb{C}P^3$ which induces an isomorphism on cohomology less than or equal to $4$.