I would highly appreciate any help you could provide. I posted this question before, now I am posting it again. Sorry for flooding SE with this question. I have computed the following things. The cohomology groups of $H^n(\mathbf{RP^2)}$ equal to
\begin{cases} \mathbf{Z} & n=0 \\ \mathbf{0} & n=1\\ \mathbf{Z_2} & n =2. \end{cases} And there are no higher-order cohomology groups.
Also, the homology groups are $H_n(\mathbf{RP^2})$ equal to \begin{cases} \mathbf{Z} & n=0 \\ \mathbf{Z_2} & n=1\\ \mathbf{0} & n\geq2. \\ \ \end{cases}
Please observe that the first order homology group is nontrivial, where the cohomology is trivial and similarly, the second-order homology group is trivial, but the cohomology group is not. My question is: What is the geometric significance/interpretation of these differences?
Most importantly, I want to know: Can it be possible that the homology and the cohomology groups are always the same for every order for some class of spaces? In that case, what can we say about the topological space? For example, $S^n$ has the same homology and cohomology groups for every order.
Please note that if the dual of a matroid is isomorphic to itself, then the rank and corank of the matroid are exactly the same. For example, if the dual of any planar graph is isomorphic to the given graph, then the size of the maximal trees is the same both for the graph and its dual. Can you please help me to gain some geometric/combinatorial intuition/motivation.
Thank you so much. Have a good one.