Homotopy in $U_{2}$ 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 $U_2$. 
