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I am reading through the proof of Borsuk - Ulam in Bredon, I have appended the proof below for reference. There are two things I don't understand in this proof:

1) In the first picture below, he writes " For any simplex $\sigma : \Delta_p \to X$, the simplex $g \circ \sigma$ is distinct from $\sigma \ldots $..... There are exactly two such liftings of the form $\sigma$ and $g \circ \sigma$." Now I understand why there are exactly two liftings but why are they of the form $\sigma$ and $g \circ \sigma$? I guess this comes to knowing why the simplices of $X$ fall into two types, which I don't understand. Is there something extra I'm not knowing?

2) I understand what the proof says, but what is boggling me is where has the assumption that $n > m$ been used? It seems to me that the exact same proof could be used to derive a contradiction if we assume $m > n$.

Thanks.

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    oh, sorry, @benjalim, it's patjennings88@gmail.com. Sorry it took so long, I was in the middle of nowhere in Bavaria. Thanks again.2012-09-17

2 Answers 2

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I have worked on 2) and this is why I think it does not work if we try to derive a contradiction from assuming $m > n$. Now in the long exact sequence above we would have

$H_m(P^n;\Bbb{Z}/2\Bbb{Z}) = \ldots = H_{n+1}(P^n;\Bbb{Z}/2\Bbb{Z}) = 0$

and no information from these extra terms is extracted. So without loss of generality we are looking at the long exact sequence

$\begin{eqnarray*} 0 &\to& H_n(P^n;\Bbb{Z}/2\Bbb{Z}) \stackrel{\cong}{\to} H_n(\Bbb{S}^n;\Bbb{Z}/2\Bbb{Z}) \stackrel{0}{\to} H_n(P^n;\Bbb{Z}/2\Bbb{Z}) \stackrel{\cong}{\to} H_{n-1}(P^n;\Bbb{Z}/2\Bbb{Z}) \to \ldots \\ \\ &&\ldots H_1(P^n;\Bbb{Z}/2\Bbb{Z}) \to \ldots \to H_0(P^n;\Bbb{Z}/2\Bbb{Z}) .\end{eqnarray*}$

However if we follow the steps in the proof as above we end up with the diagram

$\begin{array}{ccc} H_n(P^m;\Bbb{Z}/2\Bbb{Z}) &\stackrel{t_\ast}{\longrightarrow}&H_n(\Bbb{S}^m,\Bbb{Z}/2\Bbb{Z}) = 0 \\ \uparrow \psi_\ast&& \uparrow \phi_\ast \\ H_n(P^n;\Bbb{Z}/2\Bbb{Z}) &\stackrel{t_\ast}{\longrightarrow} & H_n(\Bbb{S}^n,\Bbb{Z}/2\Bbb{Z})\end{array}$

with the arrows on the left column and bottom row isomorphisms. But this tells us nothing because going round one way and going round the other we both get zero.

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    For 1) Yes, if you specialize to the covering map of the sphere (you need to be a bit careful about what you do at the equator). I don't understand your explanation of 2) in the comment.2012-09-12
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Despite somewhat rusty homology knowledge, I think I can follow the proof.

1) Just to repeat my comment on this, there's actually not much too it. If $\tau$ is a simplex in $Y$ and $\sigma$ is a lifting of $\tau$ to $X$, then $g\circ\sigma$ is another lifting of $\tau$.

2) We need some slight prior knowledge: for any $p>0$, $ H_p(S^p,\mathbb{Z}_2)\approx H_0(S^p,\mathbb{Z}_2)\approx\mathbb{Z}_2 \textrm{ while } H_q(S^p,\mathbb{Z}_2)\approx0\textrm{ for }q=1,\ldots,p-1. $

2a) The first step is to prove that $H_q(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$ for $q=0,\ldots,p$. To do this, we use the long exact sequence $ \cdots \rightarrow H_q(P^p,\mathbb{Z}_2) \rightarrow H_q(S^p,\mathbb{Z}_2) \rightarrow H_q(P^p,\mathbb{Z}_2) \rightarrow H_{q-1}(P^p,\mathbb{Z}_2) \rightarrow H_{q-1}(S^p,\mathbb{Z}_2) \rightarrow\cdots $ where $H_q(P^p,\mathbb{Z}_2)\rightarrow H_q(S^p,\mathbb{Z}_2)$ maps a simplex to the sum of both liftings of the simplex, while $H_q(S^p,\mathbb{Z}_2)\rightarrow H_q(P^p,\mathbb{Z}_2)$ just maps a simplex to the image in the ordinary way. If $2, we get $H_q(P^p,\mathbb{Z}_2)\approx H_{q-1}(P^p,\mathbb{Z}_2)$ since $H_q(S^p,\mathbb{Z}_2)\approx H_{q-1}(S^p,\mathbb{Z}_2)\approx0$.

At the lower end of the sequence we get $ 0 \rightarrow H_1(P^p,\mathbb{Z}_2) \rightarrow H_0(P^p,\mathbb{Z}_2) \rightarrow H_0(S^p,\mathbb{Z}_2) \rightarrow H_0(P^p,\mathbb{Z}_2) \rightarrow 0 $ where the first and the last maps are isomorphisms, and get $H_1(P^p,\mathbb{Z}_2)\approx H_0(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$.

At the upper end of the sequence, we get $ 0\rightarrow H_p(P^p,\mathbb{Z}_2) \rightarrow H_p(S^p,\mathbb{Z}_2) \rightarrow H_p(P^p,\mathbb{Z}_2) \rightarrow H_{p-1}(P^p,\mathbb{Z}_2) \rightarrow 0 $ where the middle map must be zero, hence $H_p(P^p,\mathbb{Z}_2)\approx H_{p-1}(P^p,\mathbb{Z}_2)\approx\mathbb{Z}_2$. The reason the middle map is zero is that the composition $H_p(S^p,\mathbb{Z}_2)\rightarrow H_p(P^p,\mathbb{Z}_2)\rightarrow H_p(S^p,\mathbb{Z}_2)$, which are first two maps of the long sequence just composed in the opposite order, has to be zero (since it maps a simplex to twice itself); but $H_p(P^p,\mathbb{Z}_2)\rightarrow H_p(S^p,\mathbb{Z}_2)$ is injective (monomorphism), so for the composition to be zero the map $H_p(S^p,\mathbb{Z}_2)\rightarrow H_p(P^p,\mathbb{Z}_2)$ must be zero.

2b) The next step is to apply the assumed equivariant map $\phi:S^n\rightarrow S^m$ for $n>m$ to obtain a contradiction.

The map $\phi_*:H_k(P^n,\mathbb{Z}_2)\rightarrow H_k(P^m,\mathbb{Z}_2)$ is an isomorphism for $k=0,\ldots,m$. This follows e.g. by induction, starting with $k=0$ and using $H_k(P^n,\mathbb{Z}_2)\approx H_{k-1}(P^n,\mathbb{Z}_2)$ and $H_k(P^m,\mathbb{Z}_2)\approx H_{k-1}(P^m,\mathbb{Z}_2)$ (commutative diagram in the book).

The final commutative diagram is now $ \begin{array}{ccc} \mathbb{Z}_2\approx H_m(P^n,\mathbb{Z}_2)&\rightarrow &H_m(S^n,\mathbb{Z}_2)\approx 0\\ \downarrow&\circ&\downarrow\\ \mathbb{Z}_2\approx H_m(P^m,\mathbb{Z}_2)&\rightarrow &H_m(S^m,\mathbb{Z}_2)\approx\mathbb{Z}_2\\ \end{array} $ where going in one direction should produce an isomorphism, while going in the other should map to zero since $H_m(S^n,\mathbb{Z}_2)\approx 0$. I.e., we have a contradiction.