A popular exercise is to show that $\mathbb{R}P^3$ is not homotopy equivalent to $\mathbb{R}P^2 \vee S^3$. The standard way is using cup products. This has been asked several times in various places and the solution exists online in several places such as here:
Show that $\mathbb{R} P^3$ is not homotopy equivalent to $\mathbb{R} P^2 \vee S^3$.
The most up-voted answer says that $H^\ast(\mathbf{RP}^2\vee S^3,\mathbf{Z}/2\mathbf{Z})=\mathbf{Z}/2\mathbf{Z}[y]/(y^3)\times\mathbf{Z}/2\mathbf{Z}[z]/(z^2)$ with $|y|=1$ and $|z|=3$
and to get this I presume they are claiming that $H^*(X \vee Y; R) \simeq H^*(X;R) \times H^*(Y;R)$. I have seen this used in other solutions also but this relationship only holds for reduced cohomology, and the reduced cohomology ring should be different than the unreduced version.
Why is it, apparently, okay to use the unreduced version?
You could simply argue using Homology. Here are some hints :
(1) Recall that the Universal cover of $\Bbb R {\Bbb P}^3$ is $S^3$.
(2) Prove that the Universal cover of $\Bbb R {\Bbb P}^2 \lor S^3$ is $S^3 \lor S^2 \lor S^3$ .
(3) Recall that $X \simeq Y \implies$ universal cover of $X \simeq $ universal cover of $Y$
(4)$H_2(S^3)=0$ whereas, $H_2(S^3 \lor S^2 \lor S^3)=\tilde{H_2}(S^2)= \Bbb Z$
Cheers!