Borsuk theorem using degree theory

Question

watching an online lecture I found the following short proof of the Borsuk theorem using degree theory.

Assuming the following statement:

Let $M^k,N^k$ be two smooth $k$-dimensional manifolds with boundary and $f:M \rightarrow N$ a continuous map, such that $f(\partial M ) \subset \partial N$ then one has $$ \operatorname{deg}_2(f)= \operatorname{deg}_2 f\vert_{\partial M} .$$ Hereby on the righthandside one considers the map $f:\partial M \rightarrow \partial N$, and $\operatorname{deg}_2$ means that one considers the degree modulo $2$.

One wants to prove the following version of the Borsuk theorem:

There is no continuous map $f:D^n \rightarrow S^{n-1}$ with $f\vert_{S^{n-1}} = id$.

The proof goes like this: Assume such a map $f$ exists by the righthandside of the argument it has degree $1$ since $f=id$ on the boundary.

Now comes the argument I do not understand: Since $f$ is not surjective the lefthandside of the statement gives that the degree of $f$ is $=0$.

Why can $f$ not be surjective? I assume that one considers the restriction of $f$ to the open ball...but still why can this map not be surjective?

Thank you very much.

Answer 1

I think it should be $f$ is homotopic to a non-surjective map $g$(constant map). And $deg_2(f)=deg_2(g)=0$. But the above theorem says $deg_2(f)=1$. SO contradiction.