If $\mathbb{Z}≅\widetilde{H_0}(\mathbb{R}P^2)⊕\mathbb{Z}$ then $\widetilde{H_0}(\mathbb{R}P^2)=0$?

Question

If I were to calculate $H_3(S^3\mathbb{R}P^2)$ where $SX$ is the suspension of the topological space $X$, then I get to $H_3(S^3\mathbb{R}P^2)=H_2(S^2\mathbb{R}P^2)=H_1(S\mathbb{R}P^2)=\widetilde{H_0}(\mathbb{R}P^2)$, so I know that $H_0(X;R)≅\widetilde{H_0}(X;R)⊕R$, so $\mathbb{Z}≅\widetilde{H_0}(\mathbb{R}P^2)⊕\mathbb{Z}$, so I can conclude then that $\widetilde{H_0}(\mathbb{R}P^2)=0$? Why? Thank you!

Answer 1

Yes you can. You can also recall the following reasoning: $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z} $ so $$ \tilde{H}_0(X) \cong \mathbb{Z}^{\#(\text{connected components of }X) - 1} $$ where if $X$ has only one connected component, then we have zero copies of $\mathbb{Z}$, i.e. $\tilde{H}_0(X) = 0$. Indeed, $\mathbb{R}P^2$ has one connected component so your result follows.

Answer 2

Let me try to explain a more general approach here. One has $${H_0}(X) \cong \mathbb{Z}^{\oplus\pi_0(X)},$$ which yields $$\widetilde{H_0}(X) \cong \mathbb{Z}^{\oplus\pi_0(X)-1}.$$ Since $\mathbb{P}^2(\mathbb{R})$ is path-connected, you get $\widetilde{H_0}(X) = 0$.

To get this result, show the special case when $X$ is path-connected and then reduce the general case to this one by using Mayer-Vietoris.