Homology of $P^n$ minus a point Denote real projective space by $P^n.$ I want to compute $H_*(P^n - p)$ (for some $p \in P^n$) given that I already know $H_*(P^n).$ Now we know that, as for any $n$-manifold, the local homology groups $H_i(P^n,P^n - p)$ at $p$ are zero except when $i = n$, in which case $H_n(P^n,P^n - p) \cong \mathbb{Z}.$ (This is seen by excising out the complement of an Euclidean ball around $p$.) So the long exact sequence of the pair $(P^n,P^n-p)$ gives $H_i(P^n,P^n-p) \cong H_i(P^n)$ when $i \neq n-1,n.$
When $n$ is odd, we have the exact sequence
$$\require{AMScd}\begin{CD}
H_{n+1}(P^n,P^n-p) @>>> H_n(P^n-p) @>>> H_n(P^n) @>>> H_n(P^n,P^n-p) @>>> H_{n-1}(P^n-p) @>>> H_{n-1}(P^n) \\
@VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V \\
0 @>>> ? @>>> \mathbb{Z} @>>> \mathbb{Z} @>>> ? @>>> 0
\end{CD}$$
The problem is that I don't understand the generators involved and what the middle map does to them. My intuition is that a generator of $H_n(P^n)$ represents a choice of orientation, a generator of $H_n(P^n,P^n-p)$ represents a choice of local orientation about $p,$ and that the middle map is an isomorphism because $P^n$ is orientable. Is this correct? If so, how does one formalize this argument? This would give that $H_n(P^n-p) = H_{n-1}(P^n-p) = 0.$
When $n$ is even, we have the exact sequence
$$\require{AMScd}\begin{CD}
H_{n+1}(P^n,P^n-p) @>>> H_n(P^n-p) @>>> H_n(P^n) @>>> H_n(P^n,P^n-p) @>>> H_{n-1}(P^n-p) @>>> H_{n-1}(P^n) @>>> H_{n-1}(P^n,P^n-p) \\
@VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V @VV{\cong}V \\
0 @>>> ? @>>> 0 @>>> \mathbb{Z} @>>> ? @>>> \mathbb{Z}/2 @>>> 0
\end{CD}$$
Thus we see that $H_n(P^n-p) = 0$ in this case. But how do we compute $H_{n-1}(P^n-p)$? Again, I'm having trouble understanding the generators and maps involved.
 A: You're working too hard. Think about the cell decomposition of $\Bbb{RP}^n$; it's obtained from $\Bbb{RP}^{n-1}$ by adding an $n$-cell by the projection $S^{n-1} \to \Bbb{RP}^{n-1}$. We may pick the point you're deleting to be in the interior of the $n$-cell (otherwise, pick a homeomorphism of $\Bbb{RP}^n$ that takes it there; the homeomorphism group of a connected manifold acts transitively).
Now, because $D^n \setminus \{0\}$ deformation retracts onto $S^{n-1}$, we see that $\Bbb{RP}^n \setminus \{p\}$ deformation retracts onto $\Bbb{RP}^{n-1}$. Because you already know the cohomology of this manifold, we're done.
A: It has already been established that this is not necessary to compute the homology of $P^n\setminus p$, but let's still have a look.
Your argument for odd $n$ is absolutely right. You should be able to find the properties that you need in any textbook that proves that for an orientable manifold $M$ you have $H_n(M;\mathbb Z)\cong\mathbb Z$ and also talks about local cohomology, say Spanier or Dold or Bredon $\texttt{;)}$
Now for even $n$ the extension can only be $\mathbb Z\xrightarrow{\cdot 2}\mathbb Z\to\mathbb Z_2$ or $\mathbb Z\to\mathbb Z\oplus\mathbb Z_2\to\mathbb Z_2$. We also know that for coefficients in $\mathbb Z_2$ the first map must actually become zero by repeating the argument for odd $n$, but now not requiring orientability. Applying a universal coefficient theorem then rules out the second case, and $H_{n-1}$ must be $\mathbb Z$.
A: The space $X=P^n\setminus p$ is homeomorphic to the total space of the tautological line bundle $\xi$ on $P^{n-1}$.
Thus $X$  is homotopically equivalent to the base space of that bundle, namely $P^{n-1}$ and we have $$  H_{n-1}(P^n\setminus p)=\mathbb Z \quad \text {if} \: n\gt 0 \: \text {is even}$$     $$ H_{n-1}(P^n\setminus p)=0 \quad \text {if} \: n\gt 1 \: \text {is odd} $$          
