Let $\mathbb{S}^{1}$ be the unit circle and $U_{2}\subset M_{2\times 2}(\mathbb{C})$ be the unitary group. Let $f,g:\mathbb{S}^{1}\rightarrow U_{2}$ be maps defined by \begin{align} f(x)= \begin{pmatrix} x & 0\\ 0 & 1 \end{pmatrix} \quad &\text{and} \quad g(x)= \begin{pmatrix} 1 & 0\\ 0 & x \end{pmatrix}. \end{align} Show that $f$ and $g$ are homotopic.
I was wondering if somebody could help me with this question. I tried to define a homotopy by using \begin{align} H(x,t)= \begin{pmatrix} (1-t)x+t & 0\\ 0 & (1-t)+tx \end{pmatrix}. \end{align} But this is not in $U_{2}$ for all $t\in [0,1]$. Also, I want to somehow use the fact that \begin{align} \begin{pmatrix} x & 0\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & x \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. \end{align}
Thanks!
Your homotopy will not work. Instead, follow the hint: $U_2$ is path-connected and the identity matrix can be connected to the matrix $$ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $$ by a certain path $M_t, t\in [0,1]$, in $U_2$.