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The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \operatorname{dim}M$, due to Schoen and Uhlenbeck [3] (see also Theorem 2.1 in [1]).

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. In fact the mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$ f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. $$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous. As $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f_\epsilon$ is a formula for the extension of the trace of the function, see [2], $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$ \left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ Since $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all $y$} $$ we have $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$ g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} $$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev spaces in mathematics. I, 185–222, Int. Math. Ser. (N. Y.), 8, Springer, New York, 2009.

[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

[3] R. Schoen, K. Uhlenbeck, Boundary regularity and the Dirichlet problem for harmonic maps. J. Differential Geom. 18 (1983), 253–268.

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \operatorname{dim}M$, due to Schoen and Uhlenbeck [3] (see also Theorem 2.1 in [1]).

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. Let $B^n=B^n(0,1)$ and let $B^n_{1+\delta}=B^n(0,1+\delta)$ for some small $\delta>0$. First of all, we can extend the mapping $f$ to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$ if $\delta>0$ is small enough. Indeed, the Sobolev extension operator $E$ is defined through averages, see [2], so extending $f$ to $B^n_{1+\delta}\setminus B^n$ gives a function $Ef\in W^{1,n}(B^n_{1+\delta},\mathbb{R}^n)$ that is continuous in the annulus $B^n_{1+\delta}\setminus B^n$. If $\delta>0$ is small enough, $|Ef|>1/2$ on $B^n_{1+\delta}\setminus B^n$ (by continuity and the fact that $|f|=1$ on $\partial B^n$) and hence $\tilde{f}=Ef/|Ef|$ in $B^n_{1+\delta}\setminus B^n$ and $\tilde{f}=f$ in $B^n$ belongs to $\tilde{f}\in W^{1,n}(B^n_{1+\delta},S^{n-1})$. If we prove that the degree of $\tilde{f}$ on the boundary of $B^n_{1+\delta}$ is zero, then also degree of $f$ on the boundary of $B^n$ is zero (by homotopy invariace of degree and continuity of $\tilde{f}$ in $B^n_{1+\delta}\setminus B^n$).

The above construction shows that we can assume that $f$ is continuous in a neighborhood of $\partial B^n$ (because $\tilde{f}$ is continuous near the boundary of the ball $B^n_{1+\delta}$ and the argument given below can be applied to $\tilde{f}$ showing that the degree of $\tilde{f}$ is zero on the boundary of the ball $B^n_{1+\delta}$).

The mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$ f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. $$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous up to the boundary: as $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f$ is continuous in an annulus near the boundary, $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$ \left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ Since $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all $y$} $$ we have $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$ g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} $$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev spaces in mathematics. I, 185–222, Int. Math. Ser. (N. Y.), 8, Springer, New York, 2009.

[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

[3] R. Schoen, K. Uhlenbeck, Boundary regularity and the Dirichlet problem for harmonic maps. J. Differential Geom. 18 (1983), 253–268.

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions.

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. In fact the mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$ f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. $$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous. As $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f_\epsilon$ is a formula for the extension of the trace of the function, see [1], $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$ \left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ Since $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all $y$} $$ we have $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$ g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} $$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

Remark. Later I will add more relevant references since the trick with the Poincare inequality used above, is due to Schoen and Uhlenbeck.

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions. The proof presented below is based on the proof of density of $C^\infty(M,N)$ in $W^{1,p}(M,N)$, $p\geq \operatorname{dim}M$, due to Schoen and Uhlenbeck [3] (see also Theorem 2.1 in [1]).

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. In fact the mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$ f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. $$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous. As $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f_\epsilon$ is a formula for the extension of the trace of the function, see [2], $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$ \left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ Since $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all $y$} $$ we have $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$ g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} $$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] P. Hajłasz, Sobolev mappings between manifolds and metric spaces. In: Sobolev spaces in mathematics. I, 185–222, Int. Math. Ser. (N. Y.), 8, Springer, New York, 2009.

[2] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

[3] R. Schoen, K. Uhlenbeck, Boundary regularity and the Dirichlet problem for harmonic maps. J. Differential Geom. 18 (1983), 253–268.

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions.

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

When I find time I will sketch a proof. I hope to do it tomorrow.

The restriction of $f$ to the boundary has degree zero. It is true also in higher dimensions.

Theorem. If $f\in W^{1,n}(B^n,S^{n-1})$ and $f|_{\partial B^n}\in C^0$, then $f|_{\partial B^n}:S^{n-1}\to S^{n-1}$ has degree zero.

Here the restriction to the boundary $f|_{\partial B^n}$ is defined as a trace of a $W^{1,n}$ function.

Proof. In fact the mapping $f$ takes values into $\mathbb{R}^{n}$ since $f:B^n\to S^{n-1}\subset\mathbb{R}^n$. Given $\epsilon>0$ define $r_{\epsilon,x}=\epsilon(1-|x|)$ and $$ f_\epsilon(x)=\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})} f(y)\, dy. $$ That is we average $f$ over a ball of radius $\epsilon$ times the distance of $x$ to the boundary of the unit ball $B^n$. The function $f_\epsilon$ is continuous. As $x$ approaches $\partial B^n$, the radius of the ball over which we average tends to zero and hence $f$ is continuous up to the boundary, because $f_\epsilon$ is a formula for the extension of the trace of the function, see [1], $f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$.

According to the Poincare inequality $$ \left(\frac{1}{|B(x,r_{\epsilon,x})|}\int_{B(x,r_{\epsilon,x})}|f(y)-f_\epsilon(x)|^n\, dy\right)^{1/n} \leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ Since $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq |f(y)-f_\epsilon(x)| \quad \text{for all $y$} $$ we have $$ \operatorname{dist}(f_\epsilon(x),S^{n-1})\leq C\left(\int_{B(x,r_{\epsilon,x})}|\nabla f|^n\right)^{1/n}. $$ The right hand side converges uniformly to $0$ in $x$. That is if $\epsilon$ is small enough, $f_\epsilon(x)\neq 0$ and hence $$ g(x)=\frac{f_\epsilon(x)}{|f_\epsilon(x)|} $$ is a continuous map $g:B\to S^{n-1}$, $g|_{\partial B^n}=f_\epsilon|_{\partial B^n}=f|_{\partial B^n}$. This shows that $\operatorname{deg}(f|_{\partial B^n})=0$.

[1] G. Leoni, A first course in Sobolev spaces. Second edition. Graduate Studies in Mathematics, 181. American Mathematical Society, Providence, RI, 2017.

Remark. Later I will add more relevant references since the trick with the Poincare inequality used above, is due to Schoen and Uhlenbeck.