Degrees of certain maps of spheres

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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How do you consider a map from $S^n$ a map from $I^n$? $S^N$ and $I^n$ are not homeomorphic... – Mariano Suárez-Alvarez Oct 29 '10 at 13:06
The 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 $(Y,y)$. – J.K.T. Oct 29 '10 at 13:10
So 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? – Mariano Suárez-Alvarez Oct 29 '10 at 13:24
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. – J.K.T. Oct 29 '10 at 14:07

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.
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.