Continuous path of Fredholm operators Let $H$ be a (infinite dimensional) Hilbert space and denote by $\mathcal{F}(H)$ the semigroup of bounded Fredholm operators in $H$. Let $S,T\in\mathcal{F}(H)$ and let $I=id_{H}$ be the identity map in $H$.
I know that $ST\oplus I$ and $S\oplus T$ are both elements of $\mathcal{F}(H\oplus H)$ and that they have the same index.

Question. How can one explicitly construct a continuous map $\alpha\colon[0,1]\to\mathcal{F}(H\oplus H)$ such that $\alpha(0) = ST\oplus I$ and $\alpha(1)=S\oplus T$ ?

Writing $A\oplus B = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$, I already notice that
$$\alpha(t) = \begin{bmatrix} ST & 0 \\ 0 & I \end{bmatrix}
\begin{bmatrix} \cos(\pi t/2) & -\sin(\pi t/2) \\ \sin(\pi t/2) & \cos(\pi t/2) \end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & S^{-1} \end{bmatrix}
\begin{bmatrix} \cos(\pi t/2) & \sin(\pi t/2) \\ -\sin(\pi t/2) & \cos(\pi t/2) \end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & S \end{bmatrix}$$
seems to do the job, though I do not know how to check that $\alpha(t)\in\mathcal{F}(H\oplus H)$ for all $t\in[0,1]$. Besides, in the above formula I've assumed that $S$ is invertible, and I could not find a solution without that assumption.
Any ideas?
Thanks in advance.
 A: The map I was looking for is $\alpha\colon[0,1]\to\mathcal{F}(H\oplus H)$ given by
$$\alpha(t) =
\begin{bmatrix} S & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} \cos(\pi t/2) & \sin(\pi t/2) \\ -\sin(\pi t/2) & \cos(\pi t/2)\end{bmatrix}
\begin{bmatrix} T& 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} \cos(\pi t/2) & -\sin(\pi t/2) \\ \sin(\pi t/2) & \cos(\pi t/2)\end{bmatrix}$$
since
$$\alpha(0) =
\begin{bmatrix} S & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} T & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} I & 0 \\ 0 & I\end{bmatrix} =
\begin{bmatrix} ST & 0 \\ 0 & I\end{bmatrix}$$
and
$$\alpha(1) =
\begin{bmatrix} S & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} 0 & I \\ -I & 0\end{bmatrix}
\begin{bmatrix} T & 0 \\ 0 & I\end{bmatrix}
\begin{bmatrix} 0 & -I \\ I & 0\end{bmatrix} =
\begin{bmatrix} S & 0 \\ 0 & T\end{bmatrix}$$
Also, $\alpha([0,1])\subseteq\mathcal{F}(H\oplus H)$ since $\begin{bmatrix} \cos(\pi t/2) \cdot I & \sin(\pi t/2)\cdot I \\ -\sin(\pi t/2)\cdot I & \cos(\pi t/2)\cdot I\end{bmatrix}$ is Fredholm in $H\oplus H$ and a composition of Fredholm operators remains Fredholm.
