I'm trying to show that $\mathbb{R}P^2 \times \mathbb{R}P^2$ and $\mathbb{R}P^4$ do not have the same Stiefel-Whitney numbers.
I have a theorem that says "If $n+1$ is not a power of 2, then $\mathbb{R}P^n$ is not a boundary of a compact manifold. Since 5 is not a power of 2, then $\mathbb{R}P^4$ is not a boundary of a compact manifold.
I also have a theorem that says "If a closed manifold is a boundary of a compact manifold, then all its Stiefel-Whitney numbers are 0."
I can't use this, correct? It's not the right logic to use the fact I have with this theorem?
I'm ultimately trying to prove that $\mathbb{R}P^2 \times \mathbb{R}P^2$ is not cobordant to $\mathbb{R}P^4$. Please help me find a way to prove this!
The SW class of $\mathbb{RP}^4$ equals $$ (1+a)^{4+1}=1+a+a^4. $$ From this it is seen that $\langle w_1^4(\mathbb{RP}^4),[\mathbb{RP^4}]\rangle=1$ and $\langle w_4(\mathbb{RP}^4),[\mathbb{RP^4}]\rangle=1$ and all other numbers vanish.
For $\mathbb{RP}^2\times \mathbb{RP}^2$ we get that the stiefel whitney class is $$ (1+a)^3(1+b)^3=1+a+b+a^2+ab+b^2+a^2b+ab^2+a^2b^2 $$ and for example $w_1^4(\mathbb{RP}^2\times \mathbb{RP}^2)=0$ so the associated Stiefel-Whitney number also vanishes.
The first Stiefel-Whitney number of $\mathbb{RP}^2\times \mathbb{RP}^2+\mathbb{RP}^4$ does not vanish, hence this manifold is not nullbordant. Thus $\mathbb{RP}^2\times \mathbb{RP}^2$ is not cobordant to $\mathbb{RP}^4$.