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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?

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    By 'not deformable' do you mean not homotopic?2012-05-27
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    Definition: Space X is deformable into a subspace A provided that there is a homotopy H:X*[0,1] to X such that H(x,0) is the identity map on X and H(x,1) is a map with range in A2012-05-27

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