Two Queries on Proof of Borsuk-Ulam in Bredon's $\textit{Geometry and Topology}$ 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.
 A: 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.
A: 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.
