0
$\begingroup$

Suppose $S^n$ is the n-sphere with basepoint $x$. If based map $f:S^n\rightarrow S^n$ is such that the pull-back $f^{-1}(S^{n}-x)$ is connected do we necessarily have that $deg(f)$ is either -1,0, or 1?

  • 0
    I think "maps of pairs" is pretty explicit and many texts define homotopy groups using them but it really doesn't matter. It has been changed.2010-10-29

2 Answers 2

3

Look at $S^2$ as the result of adding $\infty$ to $\mathbb C$, and let $f:S^2\to S^2$ be the map $z\mapsto z^2$. Then the preimage of $S^2-z$ is connected for all $z\in S^2$ (because the preimage of $z$ is finite). The degree is, of course, 2.

0

More generally I gather the suspension of a map on $S^n$ of degree $k$ is a map of degree $k$ in $S^{n+1}$. Note that if $x$ is chosen to be one of the north or south poles of $S^{n+1}$ and $|k|> 1$, then $ f^{-1}(S^{n+1}-{x})$ will always be $S^{n+1}-{x}$ which will be connected and the map will have degree $k$. You will have many counterexamples :) This is based on some stuff like maps of non-zero degree are onto.