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\begin{document}

\def\U{{\cal U}}
\def\a{\alpha}
\def\s{\sigma}
\def\d{\partial}
\def\D{\Delta^n}
\def\r{\rightarrow} 
\def\i{\hookrightarrow}
\def\l{\lambda}


{\bf Degree}

Let $f:  S^n \r S^n$ for $n > 0$.

Then $f_*:  H_n(S^n) = \Z \r \Z = H_n(S^n)$.

$f_*$ is a homomorphism and thus $f_*(\a) = d\a$.

Defn:  The  {\it degree of $f$} is $d$.

a.)  $deg ~id = 1$

b.)  $f$ not onto implies $deg~f =0$ 

Suppose $x_0 \in S^n - f(S^n)$.  Then $S^n \rightarrow S^n - \{x_0\} \i S^n$ implies $f_* = 0$ since $H_n(S^n - \{x_0\} = 0$ 

c.)  If $f$ is homotopic to $g$, then $f* = g*$ and thus $deg~f = deg~g$. 

Hopf Thm (cor 4.25):    If  $deg~f = deg~g$, then $f$ is homotopic to $g$.

 d.)  $(f\circ g)_* = f_* \circ g_*$, and thus  $deg~(f\circ g) = (deg~f)(deg~g)$ 
 
 e.) Let $S^n = \{x \in R^{n+1} ~|~   ||x|| = 1\}$.   $deg~r_i = -1$ where
  
 $r_i(x_1, ..., x_{i-1}, x_i, x_{i+1}, ..., x_{n+1}) = (x_1, ..., x_{i-1},- x_i, x_{i+1}, ..., x_{n+1})$.  
  
$S^n = \Delta_1^n ~\bigcup\limits_\partial~  \Delta_2^n$, ~~
 $H_n(S^n) = < \Delta_1^n -  \Delta_2^n>$  
 \vskip 10pt
 \rightline{and $f(\Delta_1^n -  \Delta_2^n) = -\Delta_1^n +  \Delta_2^n$}
  
  \eject
  
 f.)   The antipodal map $-id:  S^n \r S^n$,~ $-id(x) = -x$
 \vskip 10pt
\rightline{   has degree $(-1)^{n+1}$  since $r_1 \circ r_2 \circ ... \circ r_{n+1} = -id$.}
 
 
 
 g.)  If $f:  S^n \r S^n$ has no fixed points, then $deg~f = (-1)^{n+1}$
 \u
 since $f$ is homotopic to $-id$ via the homotopy
 \u\u\u
 $$F(x, t) = \frac{(1 - t)f(x) - tx}{||(1 - t) f(x)  - tx||}$$
 
If $(1 - t)f(x) - tx = 0$, then $f(x) = (\frac{t}{1-t}) x$

$x, f(x) \in S^n$ implies  $\frac{t}{1-t} = 1, -1$. 

But if $f(x) = -x$, then $(1 - t) f(x)  - tx = (1 - t) (-x)  - tx = -x$.

Thus  $(1 - t)f(x) - tx = 0$ iff $f$ has a fixed point and thus $F$ is well-defined if $f$ has no fixed points.
 
%   But if  $\frac{t}{1-t} = -1$, then $t = t -1$
h.) If $S f : S^{n+1} \r S^{n+1}$, $S([x, t]) = S([f(x), t])$ denotes
\vskip 9pt
 \rightline{ the suspension map of  $f : S^n \r S^n$, then $\deg S f = \deg f$.}
 
 The cone of of $S^n  = CS^n = (S^n \times I)/(S^n \times 1)$
 
  \vskip 9pt
 \rightline{ with base $S^n = S^n \times 0 \subset CS^n$.}
 
 $S^{n+1} = $ the suspension $SS^n = CS^n/S^n$
 
 $ H_{n+1}(CS^n) \r H_{n+1}(CS^n, S^n) \xrightarrow{\partial_*} H_{n}(S^n) \r H_{n}(CS^n) $
 
 
 i.) $f:  S^1 \r S^1$, $f(z) = z^k$ has degree $k$. 
   \vskip 9pt
 \rightline{   Thus  $S^{n-1}f:  S^n \r S^n$ has degree $k$}
   


   
         
 
 \eject
  
  
  Suppose $f:  S^n \r S^n$ and $\exists y$ such that $f^{-1}(y) = \{x_1, ..., x_m\}$. 
  
    Choose $U_l, ~V$ open such that $x_l \in U_l$,  $y \in V$, $f(U_l) \subset V$.  
    
    Then $f(U_l - x_l) \subset V - y$ and the following diagram commutes:
  
  
{$\commdiag{
 & & H_n(U_l, U_l - x_l)  &         \xrightarrow{~f_*~}                   &    H_n(V, V - y)    \cr 
 & \arrow(-3,-2) & \mapdown^{i_{U_l*}} & & \mapdown^{\cong} \cr
 H_n(S^n, S^n - x_l) & \xleftarrow{~i_*~} & H_n(S^n, S^n - f^{-1}(y))  &         \xrightarrow{~f_*~}                   &    H_n(S^n, S^n - y)    \cr 
 & \arrow(-3,2) & \mapup^{j} & & \mapup^{\cong} \cr
  & & H_n(S^n)  &         \xrightarrow{~f_*~}                   &    H_n(S^n)    \cr 
}$}


$f_*: H_n(U_l, U_l - x_l) = \Z \r \Z = H_n(V, V - y)$, $f_*(\a) = d_l\a$.

Defn:  The {\it local degree} of $f$ at $x_l$ = $deg~f|_{x_l} = d_l$.



Prop:  $deg~f =  \sum\limits_{l = 1}^m deg~f|_{x_l}$

 $H_n(S^n,~ S^n - f^{-1}(y)) \cong  H_n(\sqcup U_l,~ \sqcup U_l - f^{-1}(y))$
 \vskip 5pt
\rightline{$   = \oplus H_n(U_l,~ U_l - x_l)   = \oplus \Z$.}

$(i_* \circ j)(1) = 1$.  Thus $j(1) = (1, 1, ..., 1) = \sum i_{U_l*}(1)$

%$f_* \circ  i_{U_l*}(1) = d_l$,  Thus  
$f_* \circ j(1) = (1, 1, ..., 1) = \sum f_* \circ i_{U_l*}(1) = \sum  d_{l}$

\vfill

Note:  If $f:  U_l \r V$ is a homeomorphism, then $deg~f|_{x_l} = \pm 1$

\eject

{Theorem 2.28:}
  A continuous nonvanishing vector field on $S^n$ exists if and only if $n$ is odd.

Proof:  ($\Rightarrow$) Suppose $\exists$ a  continuous nonvanishing vector field, $v$, on $S^n$

Normalize the vector field so that $|v(x)| = 1$ for all $x$.  

Then $v(x) \in S^n$ and $v(x)$ is perpendicular to $x$.

Thus $ (cos(\pi t))x + (sin(\pi t)) v(x) \in S^n$.

Then $F(x, t) = (cos(\pi t))x + (sin(\pi t)) v(x)$ is a homotopy between the identity map on $S^n$ and the antipodal map.  

Thus $1 = (-1)^{n+1}$ and $n$ is odd.

 ($\Leftarrow$)  Let  $v(x_1, x_2, ..., x_{2l - 1}, x_{2l}) = (-x_2, x_1, ..., -x_{2l}, x_{2l-1})$


\vfill
\hr

{Proposition 2.29:}
  If $n$ is even, then $\Z_2$ is the only nontrivial group that can act freely on $S^n$.

Suppose $G$ acts on $S^n$.  Then $g \in G$  defines a homeomorphism $g:  S^n \r S^n$.    Since $g$ is a homeomorphism, $deg ~g = \pm 1$.  


$d:  G \rightarrow \{\pm 1\}$, $d(g) = deg~g$ is a homomorphism by 
property d.

If the action is free, then if $g \not = e$, $d(g) = (-1)^{n+1}$ by property g.

Thus if $n$ is even, $g \not = e$ implies $d(g) = -1$ , Thus $ker(d) = e$ and $d$ is an isomorphism.  Thus $G \cong  \{\pm 1\} \cong \Z_2$






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