# Show that if $f:S^8 \to S^8$ and $\|f(x) + x\| < 1$ for all $x$ then $f$ is not homotopic to the identity map

I need to show that if a map $f:S^8 \to S^8$ satisfies $\|f(x) + x\| < 1$ for all $x \in S^8$ then $f$ is not homotopic to the identity map.

### Progress

I saw some two related propositions:

• Proposition 1: Antipodal map $f: S^{n} \rightarrow S^{n}$ has degree $\deg(f) = (-1)^{n+1}$.
• Proposition 2: If $n$ is even than $f$ is not homotopic to identity.

Is this related? I also not sure, how can I show that $f$ in my question is homotopic to antipodal map?

• Show it's homotopic to $g(x)=-x$. – user147263 Mar 7 '15 at 20:11
• I saw some two related propositions: Proposition 1: Antipodal map $f: S^{n} \rightarrow S^{n}$ has degree $deg(f) = (-1)^{n+1}$. Proposition 2: If $n$ is even than $f$ is not homotopic to identity. Is this related? I also not sure, how can I show that $f$ in my question is homotopic to antipodal map? – cactus Mar 8 '15 at 12:20
• Use the straight-line homotopy between $f$ and the antipodal map $g$. By which I mean, take convex combination of $f$ and $g$ and normalize. – user147263 Mar 8 '15 at 16:30