Compute the homology groups $H_∗(X, X−p)$ for points $p∈X$ where $X = S(\Bbb CP^2)$

Question

Let $X = S(\Bbb CP^2)$ be the suspension of $\Bbb CP^2$ . Compute the homology groups $H_∗(X, X−p)$ for points $p∈X$ .

My attempt :

For any topological space $Y$ , $S(Y)-pt \simeq C(Y)$ where $C(Y)$ is the cone of Y . Thus,

$H_∗(S(\Bbb CP^2),S(\Bbb CP^2) −pt)\cong H_∗(S(\Bbb CP^2),C(\Bbb CP^2)\cong \tilde{H_∗}(S(\Bbb CP^2)/ C(\Bbb CP^2))$

Now, I haven't proved it, but seems intuitively true that for any $Y$ $S(Y)/C(Y) \cong C(Y)/Y\cong S(Y)\ldots(*)$

If $(*)$ is true , $H_∗(S(\Bbb CP^2),S(\Bbb CP^2) −pt)\cong \tilde{H_∗}(S(\Bbb CP^2)/ C(\Bbb CP^2)) \cong \tilde{H_∗}(S(\Bbb CP^2))$

Then we get that $ \tilde{H_∗}(S(\Bbb CP^2)) \cong \tilde{H_{∗-1}}(\Bbb CP^2) = \Bbb Z \text { if }* =3,5 , \text{ and } = 0 \text{ otherwise } $

So are the homology groups $H_∗(X, X−p)= \Bbb Z \text { if }* =3,5 , \text{ and } = 0 \text{ otherwise } $ ?

Answer 1

For the trivial part i.e. suspension points what you did is fine.

Suppose $p$ is not a suspension point. Then $$\tilde H_* (S(\mathbb C \mathbb P^2, S(\mathbb C \mathbb P^2)-p ) = \tilde H_* ( \mathbb C\mathbb P^2 \times (-1,1) , \mathbb C \mathbb P^2 \times (-1,1) -\{ p \} )$$ by excision

Then we have again $p$ has a neighbourhood in $CP^2 \times (-1,1)$ homeomorphic to $\mathbb R^5$. Thus another application of excision gives you it is equal to

$$\tilde H_* (\mathbb R^5, \mathbb R^5 -0)$$