Let $\mathbb{P}^2$ denote the projective plane.
Given (no need to prove) that $H_1(\mathbb{P}^2) \cong \mathbb{Z}_2$ ,$H_2(\mathbb{P}^2) \cong 0$ and the open Möbius band $M$ is homotopy equivalent to $\mathbb{S}^1$.
a), Show $\mathbb{P}^2$ is not deformable into $M$.
b), Is $\mathbb{P}^2$ deformable into any proper subspace?
My approach for a): If one assumes $\mathbb{P}^2$ is deformable into $M$, then the short exact sequence $H_2(S^1) \overset{i_{*}} \to H_2(\mathbb{P}^2) \overset{j_{*}} \to H_2(\mathbb{P}^2,S^1) \overset{\partial} \to H_1(S^1)$ has the property that $i_{*}$ is epimorphism, $j_{*}$ is trivial and $\partial$ is monomorphism, moreover one has $H_1(M) \cong H_1(\mathbb{P}^2) \oplus H_2(\mathbb{P}_2,M)$ which is equivalent to $\mathbb{Z} \cong \mathbb{Z}_2 \oplus H_2(\mathbb{P}_2,M)$. My aim for the next step is to show $H_2(\mathbb{P}_2,M)$ has finite order for contradiction to the assumption, is there anyway to achieve this by the exact sequence?