Retraction and deformation of P2 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?
 A: I'm not so sure I've completely understood you proof since it seems you skipped some logical steps.
Anyway for point a) based on your definition of deformation, if $\mathbb P^2$ deforms on the Möbius band $M$ then there should be an homotopy between $1_{\mathbb P^2}$ and a mapping of the form $i \circ r \colon \mathbb P^2 \to \mathbb P^2$ where $i \colon M \to \mathbb P^2$ is the embedding of the Möbius band in $\mathbb P^2$ and $r \colon \mathbb P^2 \to M$ is a map sending the projective real plane in the Möbius band. This is equivalent to require that the embedding $i$ has a right inverse and so by functoriality of homology we should have
$$H_n(i) \circ H_n(r) = H_n(i \circ r) = H_n(1_{\mathbb{P}^2})=1_{H_n(\mathbb P^2)}$$
but the morphism
$$H_1(r) \colon H_1(\mathbb P^2) \cong \mathbb Z_2 \to H_1(M) \cong \mathbb Z$$
so it must be the null homomorphisms, since it should send the generator of $\mathbb Z_2$ in an element of finite order in $\mathbb Z$ which can be just $0$.
From this should follows that
$$1_{H_1(\mathbb P^2)} = H_1(i \circ r) = 0$$
which cannot be possible since $H_1(\mathbb P^2) \ne 0$.
So we must conclude that there cannot be any deformation of $\mathbb P^2$ on $M$.
About point b) I'm not sure at the time, but I gonna think about.
