I think this problem is basic, and possibly it has already been discussed. I would highly appreciate if someone could spend some time, and write some easy answer without using any advanced topology concept. I used the "Universal Coefficient Theorem" to compute the cohomology groups and found that $H^n(\mathbf{RP^2)}$ equals to
\begin{cases} \mathbf{Z} & n=0 \\ \mathbf{0} & n=1\\ \mathbf{Z_2} & n =2 \end{cases} There are no higher order cohomology groups.
Note that the homology groups are the following: $H_n(\mathbf{RP^2})$ equals to \begin{cases} \mathbf{Z} & n=0 \\ \mathbf{Z_2} & n=1\\ \mathbf{0} & n\geq2 \\ \ \end{cases}
I have 4 questions. (1) Is my computation correct? (2) The first order homology group is nontrivial, where the cohomology is trivial. What is the geometric interpretation of this result? (3) Similarly, the second-order homology group is trivial, but the cohomology group is not. what is the geometric significance of these differences? (4) Most importantly, is it possible that homology and cohomology groups are always the same for every order? 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. What are classes of such spaces where we have the same homology and cohomology groups?
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. I am looking for this kind of geometric intuition/motivation.
Thank you so much for your time.