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?
Degrees of certain maps of spheres
0
$\begingroup$
algebraic-topology
-
0*How* do you consider a map from $S^n$ a map from $I^n$? $S^N$ and $I^n$ are not homeomorphic... – 2010-10-29
-
0The space of relative maps $Map((I^n,\partial I^n),(Y,y))$ (i.e. maps $f:I^n\rightarrow Y$ such that $f(\partial I^n)=y$) is naturally homeomorphic to the based mapping space $Map((S^n,x),(Y,y))$ for any based space $(Y,y)$. – 2010-10-29
-
0So you are identifying $f$ with its composition with an identification map $I^n\to S^n$ which collapses the boundary to a point and nothing else. Why not make this explicit? – 2010-10-29
-
0I 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
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.