Let $f : \mathbb{R}P^2 \to \mathbb{S}^2$ be the following continuous map: choose a point $p \in \mathbb{S}^2$,put $f(\mathbb{R}P^1) =\{p\}$, and let $f$ map the disc $\mathbb{R}P^2 \setminus \mathbb{R}P^1$ homeomorphically onto the disc $\mathbb{S}^2 \setminus \{p\}$.
Notice that $f: (\mathbb{R}P^2, \mathbb{R}P^1) \to (\mathbb{S}^2,\{p\})$ is a relative homeomorphism. Since the spaces are good, I know that $f_* : H_2(\mathbb{R}P^2, \mathbb{R}P^1) \to H_2(\mathbb{S}^2, \{p\})$ is an isomorphism.
How do I show that $f^* : H^2(\mathbb{S}^2, \{p\}) \to H^2(\mathbb{R}P^2, \mathbb{R}P^1)$ is an isomorphism?
The argument I am thinking of works for both homology and cohomology (in all degrees):
The map $q^*:H^n(\mathbb{R}P^2/\mathbb{R}P^1,\mathbb{R}P^1/\mathbb{R}P^1) \rightarrow H^n (\mathbb{R}P^2,\mathbb{R}P^1)$ is an isomorphism since our pairs are good. It is also natural in the sense that if $f':(\mathbb{R}P^2/\mathbb{R}P^1,\mathbb{R}P^1/\mathbb{R}P^1) \rightarrow (S^2,*)$ is the induced quotient map then $f^* q^* = q^* f'^*$. Since $q^*$ is an isomorphism and $f'$ is a relative homeomorphism, we have that $f^*$ is an isomorphism.