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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? –  Juan S May 27 '12 at 8:06
    
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 A –  user31899 May 27 '12 at 23:58

1 Answer 1

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.

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