prove that elements in $K_1(A)$ coincide Let $A$ be a unital $C^*$-algebra $u\in A$ unitary and $s\in A$ isometry.
I already proved that $sus^*+(1-ss^*)$ is an unitary. 

Why is $[u]_1=[sus^*+(1-ss^*)]_1\in K_1(A)$?



*

*Basic definitions:


For $A$ unital it is $K_1(A)=U_{\infty}(A)/\sim_1$, where $U_\infty(A)=\bigcup\limits_{k=1}^\infty U(M_k(A))$, $U(M_k(A))$ denotes the group of unitary elements of $M_k(A)$, and $\sim_1$ on $U_{\infty}(A)$ is defined as follows:
For $u\in U(M_n(A)), v\in U(M_m(A))$ write $u\sim_1 v$ if there exists a natural number $k\ge max\{n,m\}$ such that $u\oplus 1_{k-n}\sim_h v\oplus 1_{k-m}$ in $U(M_k(A))$ (for more details see "An introduction to K-theory for $C^*$-algebras" ). With $[u]_1$ we denote the equivalence class of $u$ in $K_1(A)$.


*What I have done so far:


Since $sus^*+(1-ss^*)$ is an unitary, it defines an element $[sus^*+(1-ss^*)]_1\in K_1(A)$. Let $v=\begin{pmatrix} s & 1-ss^* \\ 0 & s^* \end{pmatrix}\in M_2(A)$. $v$ is an unitary and satisfies $$v\begin{pmatrix} u & 0 \\ 0 & 1 \end{pmatrix}v^*=\begin{pmatrix} sus^*+(1-ss^*) & 0 \\ 0 & 1 \end{pmatrix},$$ i.e. $v(u\oplus 1_A) v^*= (sus^*+(1-ss^*))\oplus 1_A$.
But how to conclude that $[u]_1=[sus^*+(1-ss^*)]_1\in K_1(A)$, how to define the homotopy?
 A: What you need is the Whitehead Lemma: If $v$ is a unitary in $A$, then
$$
\begin{pmatrix} v & 0 \\ 0 & v^{\ast} \end{pmatrix} \sim_h \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ in } U(M_2(A))
$$
which itself follows from the fact that
$$
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \sim_h \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ in } U(M_2(\mathbb{C}))
$$
Now, for your problem, you want to show is that if $u_1, u_2\in U(A)$ are two unitaries such that $u_1 = vu_2v^{\ast}$ for some unitary $v\in A$, then $[u_1]_1 = [u_2]_1$. For this, consider
$$
\begin{pmatrix} v & 0 \\ 0 & v^{\ast} \end{pmatrix}\begin{pmatrix} u_2 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} v^{\ast} & 0 \\ 0 & v \end{pmatrix} = \begin{pmatrix} v & 0 \\ 0 & v^{\ast} \end{pmatrix}\begin{pmatrix} u_2v^{\ast} & 0 \\ 0 & v \end{pmatrix} = \begin{pmatrix} u_1 & 0 \\ 0 & 1\end{pmatrix}
$$
So by Whitehead's Lemma, you have
$$
u_2\oplus 1 \sim_h u_1\oplus 1
$$
which proves that $[u_2]_1 = [u_1]_1$ in $K_1(A)$.
In other words, in your case, while the homotopy may not exist in $U(M_2(A))$, it does exist in $U(M_4(A))$.
