I'm studying for my topology qualifying exam, and I'm having trouble computing the homology of the connected sum of $\mathbb{R}P^4$ and $\mathbb{C}P^2$. I tried using a relative long exact sequence and Mayer-Vietoris, and I've gotten closer with the relative sequence. Here's what I have so far:
Let $X$ be the connected sum $A$ be the $S^3$ subspace of $X$ that joins $\mathbb{R}P^4$ and $\mathbb{C}P^2$. Then, $(X,A)$ is a good pair so $H_n(X,A)\cong H_n(X/A)$ and $X/A$ is $\mathbb{R}P^4\vee\mathbb{C}P^2$. The long exact sequence we get is $$\dots\to H_4(A)\to H_4(X)\to H_4(X/A)\to H_3(A)\to H_3(X)\to H_3(X/A) \to H_2(A)\to H_2(X)\to H_2(X/A)\to H_1(A)\to H_1(X)\to H_1(X/A)\to \tilde{H}_0(A)$$ $$\dots\to 0\to H_4(X) \to \mathbb{Z} \to \mathbb{Z} \to H_3(X)\to \mathbb{Z}/2\mathbb{Z} \to 0\to H_2(X)\to \mathbb{Z} \to 0\to H_1(X)\to \mathbb{Z}/2\mathbb{Z}\to 0$$
By exactness, we get that $H_1(X)\cong\mathbb{Z}/2\mathbb{Z}$ and $H_2(X)\cong\mathbb{Z}$ and since $X$ is nonorientable, we know $H_4(X)\cong 0$ but I'm having trouble figuring out what $H_3(X)$ is. It seems like it's either $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}$ depending on what these maps are. I don't have much intuition about what the maps are when we're looking at the higher dimensions. Can anyone lead me in the right direction?