We know $H^d(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2$. So there are two classes of $\mathbb{Z}_2$ generators, trivial and nontrivial, for $d=0,1,2,3,4,5$.

Wha are the Poincaré dual $(5-d)$-dimensional manifolds of the generators PD($a^d$) of $a^d\in H^d(\mathbb{RP}^5,\mathbb{Z}_2)$, for $d=0,1,2,3,4,5$?

\begin{array}{|c|c|} \hline \text{PD}(a^d)& a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2) &a^d \neq 0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2)\\ \hline d=0 &?& \mathbb{RP}^5?\\ \hline d=1 &? &\mathbb{RP}^4?\\ \hline d=2 &?& \mathbb{RP}^3?\\ \hline d=3 &?&\mathbb{RP}^2?\\ \hline d=4 &?&\mathbb{RP}^1?\\ \hline d=5 &?&\mathbb{RP}^0=\text{a point}?\\ \hline \end{array}

• It is said that if $a^d=0$. A Poincaré dual for $a\in H^d(M;\mathbb{Z}/2)$ is any embedded closed $(5-d)$-manifold which bounds, such as a small $S^{5-d}$ around a point, and this is orientable. Is this a correct generator for the trivial class $a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2)$?

We know $H^d(\mathbb{RP}^5,\mathbb{Z}_2)=\mathbb{Z}_2$. So there are two classes of $\mathbb{Z}_2$ generators, trivial and nontrivial, for $d=0,1,2,3,4,5$.

Wha are the Poincaré dual $(5-d)$-dimensional manifolds of the generators PD($a^d$) of $a^d\in H^d(\mathbb{RP}^5,\mathbb{Z}_2)$, for $d=0,1,2,3,4,5$?

\begin{array}{|c|c|} \hline \text{PD}(a^d)& a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2) &a^d \neq 0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2)\\ \hline d=0 &?& \mathbb{RP}^5?\\ \hline d=1 &? &\mathbb{RP}^4?\\ \hline d=2 &?& \mathbb{RP}^3?\\ \hline d=3 &?&\mathbb{RP}^2?\\ \hline d=4 &?&\mathbb{RP}^1?\\ \hline d=5 &?&\mathbb{RP}^0=\text{a point}?\\ \hline \end{array}

• It is said that if $a^d=0$, a Poincaré dual for $a\in H^d(M;\mathbb{Z}/2)$ is any embedded closed $(5-d)$-manifold which bounds, such as a small $S^{5-d}$ around a point, and this is orientable. Is this a correct generator for the trivial class $a^d =0 \in H^d(\mathbb{RP}^5,\mathbb{Z}_2)$?