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?
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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. |
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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. |
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