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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?

enter image description here

2) I have appended a proof of Theorem 20.1 of Bredon below. It is in two pictures. Now 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.

enter image description here enter image description here

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If $\tau$ is a simplex in $Y$ and $\sigma$ is a lifting of $\tau$ to $X$, then $g\circ\sigma$ is another lifting of $\tau$. Don't think there's any more to it than that. –  Einar Rødland Sep 12 '12 at 0:57
    
@EinarRødland Thanks, I think I'm overcomplicating 1) above. How about 2)? I have posted an answer below trying to reason out where the proof in Bredon uses the assumption that $n>m$. Is my answer below correct? –  fpqc Sep 12 '12 at 1:01
    
Sorry, but not only is my homology rusty after years of disuse, I was never used to doing it over $\mathbb{Z}_2$. Maybe, if my mind clears up a little, I'll see it. However, I note that it suffices to prove the results for $n=m+1$ since the result will always apply to an $S^{m+1}\subset S^n$. –  Einar Rødland Sep 12 '12 at 15:41
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You have a pdf copy of Brendon? I can't believe I paid so much for the hard cover now. –  Holdsworth88 Sep 12 '12 at 22:32
    
@Holdsworth88 Shhhhh...........I can send you a pdf copy if you would like one :D –  fpqc Sep 12 '12 at 22:47
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2 Answers

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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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This is correct. For 1) just note that you can choose a fundamental domain for the action of $\mathbb{Z}/2\mathbb{Z}$ given by $g$ on $X$ which gives the desired decomposition. For 2) it might be interesting to exhibit an equivariant map $S^n \to S^m$ for $n \leq m$... –  t.b. Sep 12 '12 at 2:06
    
@t.b. Thanks Theo. For 1), I can choose my fundamental domain to be the upper hemisphere yes? For 2) say I look at the case $S^2$ and $S^1$. Now let $S^1$ be the equator on the $x-y$ plane. Then given any point $a$ on $S^2$, I can draw a verticle "line" down that intersects the equator, and slide my point $a$ down that line so it is sent to the equator. So in summary, my understanding of why the proof won't give a contradiction assuming $m > n$ is correct? Thanks. –  fpqc Sep 12 '12 at 2:11
    
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. –  t.b. Sep 12 '12 at 2:23
    
@t.b. The map I wrote above is not well-defined. –  fpqc Sep 12 '12 at 3:21
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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 $2<q<p$, 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.

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