homology of $ℝ{\mathbb{P}}^3$.

We need for this problem knowledge of the homology groups of $S^2$ and $ℝ{\mathbb{P}}^2$. We will simply assume the former:

$H_k(S^2;\mathbb{Z})$

Now, for $ℝ{\mathbb{P}}^2$, we can argue without Mayer-Vietoris. $X=ℝ{\mathbb{P}}^2$ is connected, so $H_0(X;\mathbb{Z})=\mathbb{Z}$. $X$ is non-orientable, so $H_2(X;\mathbb{Z})$ is 0. Last, $H_1(X;\mathbb{Z})$ is the abelianization of the already abelian fundamental group ${\pi }_1(X)=\mathbb{Z}/2\mathbb{Z}$, so we have:

$H_k(ℝ{\mathbb{P}}^2;\mathbb{Z})$

Now that we have the homology of $ℝ{\mathbb{P}}^2$, we can compute thehomology of $ℝ{\mathbb{P}}^3$ from Mayer-Vietoris. Let $X=ℝ{\mathbb{P}}^3$,$V=ℝ{\mathbb{P}}^3\backslash \backslash \{pt\}\sim ℝ{\mathbb{P}}^2$ (by vieweing $ℝ{\mathbb{P}}^3$ as a CW-complex), $U\sim D^3\sim \{pt\}$, and $U\cap V\sim S^2$, where $\sim$ denotes equivalence through a deformation retract. Then the Mayer-Vietoris sequence gives