If a map from an n-sphere to itself has no fixed points, it's degree is $(-1)^n+1$

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This is an exercise problem in Bredons book "Topology and Geometry"



If $f:S^n rightarrow S^n$ has no fixed points, $(-f)(x)neq -x$ for all $x$.
Thus $(-f)simeq -1:S^nrightarrow S^n:xrightarrow-x$.



This implies that their degrees are equal:



$deg((-1)circ f) = deg(-1)=deg(-1)*deg(f)$



since $deg(-1) neq 0$ we conclude that $deg(f) = 1$.



Where did I make a mistake?




general-topology algebraic-topology




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  • 2




    Please explain your second sentence ("Thus $-fsimeq -1$.").
    – Ted Shifrin
    Jul 24 at 15:46











  • Ah, I realize my mistake now, $-f simeq id$ and the rest follows
    – Noel Lundström
    Jul 24 at 16:00










  • There you go. :)
    – Ted Shifrin
    Jul 24 at 16:02














up vote
0
down vote

favorite












This is an exercise problem in Bredons book "Topology and Geometry"



If $f:S^n rightarrow S^n$ has no fixed points, $(-f)(x)neq -x$ for all $x$.
Thus $(-f)simeq -1:S^nrightarrow S^n:xrightarrow-x$.



This implies that their degrees are equal:



$deg((-1)circ f) = deg(-1)=deg(-1)*deg(f)$



since $deg(-1) neq 0$ we conclude that $deg(f) = 1$.



Where did I make a mistake?




general-topology algebraic-topology




share|cite|improve this question

















  • 2




    Please explain your second sentence ("Thus $-fsimeq -1$.").
    – Ted Shifrin
    Jul 24 at 15:46











  • Ah, I realize my mistake now, $-f simeq id$ and the rest follows
    – Noel Lundström
    Jul 24 at 16:00










  • There you go. :)
    – Ted Shifrin
    Jul 24 at 16:02












up vote
0
down vote

favorite









up vote
0
down vote

favorite











This is an exercise problem in Bredons book "Topology and Geometry"



If $f:S^n rightarrow S^n$ has no fixed points, $(-f)(x)neq -x$ for all $x$.
Thus $(-f)simeq -1:S^nrightarrow S^n:xrightarrow-x$.



This implies that their degrees are equal:



$deg((-1)circ f) = deg(-1)=deg(-1)*deg(f)$



since $deg(-1) neq 0$ we conclude that $deg(f) = 1$.



Where did I make a mistake?




general-topology algebraic-topology




share|cite|improve this question













This is an exercise problem in Bredons book "Topology and Geometry"



If $f:S^n rightarrow S^n$ has no fixed points, $(-f)(x)neq -x$ for all $x$.
Thus $(-f)simeq -1:S^nrightarrow S^n:xrightarrow-x$.



This implies that their degrees are equal:



$deg((-1)circ f) = deg(-1)=deg(-1)*deg(f)$



since $deg(-1) neq 0$ we conclude that $deg(f) = 1$.



Where did I make a mistake?





general-topology algebraic-topology





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 15:46
























asked Jul 24 at 15:38









Noel Lundström

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  • 2




    Please explain your second sentence ("Thus $-fsimeq -1$.").
    – Ted Shifrin
    Jul 24 at 15:46











  • Ah, I realize my mistake now, $-f simeq id$ and the rest follows
    – Noel Lundström
    Jul 24 at 16:00










  • There you go. :)
    – Ted Shifrin
    Jul 24 at 16:02












  • 2




    Please explain your second sentence ("Thus $-fsimeq -1$.").
    – Ted Shifrin
    Jul 24 at 15:46











  • Ah, I realize my mistake now, $-f simeq id$ and the rest follows
    – Noel Lundström
    Jul 24 at 16:00










  • There you go. :)
    – Ted Shifrin
    Jul 24 at 16:02







2




2




Please explain your second sentence ("Thus $-fsimeq -1$.").
– Ted Shifrin
Jul 24 at 15:46





Please explain your second sentence ("Thus $-fsimeq -1$.").
– Ted Shifrin
Jul 24 at 15:46













Ah, I realize my mistake now, $-f simeq id$ and the rest follows
– Noel Lundström
Jul 24 at 16:00




Ah, I realize my mistake now, $-f simeq id$ and the rest follows
– Noel Lundström
Jul 24 at 16:00












There you go. :)
– Ted Shifrin
Jul 24 at 16:02




There you go. :)
– Ted Shifrin
Jul 24 at 16:02















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