It is easy to see that if a map $f:X \rightarrow Y$ is surjective then its suspension $Sf:SX \rightarrow SY$ is also surjective, and since suspension preserves degree it suffices to construct a map $f:S^{1} \rightarrow S^{1}$ of degree zero and apply repeated suspension to get the maps for $n > 1$.
I am thinking of $S^{1}$ as lying in $\mathbb{R}^{2}$ and first projecting it onto the $x$-axis followed by identifying the boundary of the of the resulting 1-dimensional disk to a point, I called these maps $p$ and $q$ respectively, so $f=pq$. I chose the point $y\in S^{1}$ which has the $f^{-1}(y)=\{x_{1},x_{2}\}$ as in my picture. The degree of $f$ is the sum of the local degrees at $x_{1}$ and $x_{2}$, and the local degree will be $\pm1$ since can choose neighborhoods smartly to get local homeomorphisms at those points.
My intuition tells me that these local degrees should have opposite signs because if we think of the circle has being oriented ccw then neighborhoods around them will project down with orientations in opposite directions.
How can I justify that their degrees are "opposite" or actually compute the degree?
Best Answer
First project $S^n$ to an interval $I$. Then compose this projection with a space-filling curve $I \to S^n$.