Winding number of a loop

Question

I am reading a paper and unfortunately I have found a problem. I have a continuos map $[0,1]\times [0,1]\to U_q(\mathbb{C})$, where $U_q(\mathbb{C})$ are unitary matrices. We can consider the map $(\lambda,\mu)\to det(Y_{\lambda,\mu})$ obtained by the restriction to the boundary of the square $[0,1]\times [0,1]$. Now the author says "Since $(\lambda,\mu)\to Y_{\lambda,\mu}$ is continuos on $[0,1]^2$, the winding number of the restriction of $(\lambda,\mu)\to det(Y_{\lambda,\mu})$ to the boundary must be zero". Unfortunately, I don't understand why the winding number is $0$. Can anyone help me? I thank you in advance for the help.

Answer 1

Let us think of $S^1$ as $S^1 = \{ z \in \mathbb{C} \, | \,|z| = 1 \}$. If you have a continuous map $f \colon S^1 \rightarrow S^1$ that extends to a map $\tilde{f} \colon D \rightarrow S^1$, where $D = \{ z \in \mathbb{C} \, | \, |z| \leq 1 \}$ is the closed unit disc, then the extension $\tilde{f}$ provides a homotopy $F_t \colon [0,1] \times S^1 \rightarrow S^1$ defined by $F_t(\theta) = \tilde{f}(te^{i\theta})$ from $F_0 \equiv \tilde{f}(0)$ to $F_1 = f$. Thus, $f$ is homotopic to a constant map and so the winding number of $f$ is zero. Alternatively, if you have a map $\tilde{f} \colon D \rightarrow S^1$, then $f(e^{i\theta})$ has winding number zero. Now replace $D$ with $[0,1]^2$.