How to show two equivalent projection in a $C^*$ algebra are not homotopic 
Show that two equivalent projections need not be homotopic. HINT: Let $P=\begin{pmatrix} 1&0\\0&0\end{pmatrix}$ and $Q=\begin{pmatrix}
        t&\sqrt{t(1-t)}\\\sqrt{t(1-t)}&1-t
    \end{pmatrix}\in M_2(C[0,1])$ be two projections. Show they're contained in a common $C^*$ subalgebra of $M_2(C[0,1])$ in which they're equivalent but not  homotopic.

My attempt: for $X=\begin{pmatrix}
        \sqrt{t}&\sqrt{1-t}\\0&0
    \end{pmatrix}$ we have that: $$XX^*=\begin{pmatrix}
        \sqrt{t}&\sqrt{1-t}\\0&0
    \end{pmatrix}\cdot \begin{pmatrix}
        \sqrt{t}&0\\\sqrt{1-t}&0
    \end{pmatrix}=\begin{pmatrix}
        1&0\\0&0
    \end{pmatrix}=P$$ and $$X^*X=\begin{pmatrix}
        \sqrt{t}&0\\\sqrt{1-t}&0
    \end{pmatrix}\cdot \begin{pmatrix}
        \sqrt{t}&\sqrt{1-t}\\0&0
    \end{pmatrix}=\begin{pmatrix}
         t&\sqrt{t(1-t)}\\\sqrt{t(1-t)}&1-t
    \end{pmatrix}=Q $$
so $P,Q\in C^*(X)$ are equivalent. However I'm not sure how to show they're not homotopic as projections in $C^*(X)$. Any help would be appreciated.
 A: Consider the Cuntz algebra $\mathcal{O}_2$ and denote the generators of $\mathcal{O}_2$ by $S_1, S_2\in\mathcal{O}_2$.
Put $p=S_1S_1^*$. Then $p$ and $1$ are equivalent, since we have $p=S_1S_1^*$ and $1=S_1^*S_1$.
However, $1$ and $p$ are not homotopic in the set of projections of $\mathcal{O}_2$. Indeed, assume that $\{p_t\}_{0\leq t\leq 1}$ is a path from $p$ to $1$. Then the value of $\lVert 1-p_t\rVert$ is $0$ or $1$, since $1-p_t$ is a projection. Since $\lVert 1-p_0\rVert=\lVert 1-p\rVert=1$, we have $\lVert 1-p_t\rVert=1$ for all $t\in[0,1]$ by the continuity of $\{p_t\}$. Therefore, we obtain $\lim_{t\to 1}p_t\not=1$,
which contradicts to the assumption that $\{p_t\}$ is a path to $1$.
I am sorry that I cannot make use of your hint.
A: Let $A$ be the subalgebra of $C([0, \pi ], \mathbb R)$ formed by the functions $f$ such that $f(0)=f(\pi)\in {\mathbb R}e_{11}$.  Consider
$$
  X(t)=\pmatrix{\cos(t) & 0 \cr \sin(t) & 0}, \quad 0\leq t\leq \pi ,
  $$
and put
$$
  P=X^*X=\pmatrix{1 & 0 \cr 0 & 0},
  $$
and
$$
  Q=XX^*=\pmatrix{\cos^2(t) & \cos(t)\sin(t) \cr \cos(t)\sin(t) & \sin^2(t)}.
  $$
Clearly $X$ lies in $A$, so both $P$ and $Q$ lie in $A$ and $P\sim Q$.  However $P$ and $Q$ are not homotopically
equivalent for the following reason.
Given any potential homotopy $P_s$ between $P$ and $Q$, every $P_s$ will necessarily satisfy
$$
  \text{rank}(P_s(t))=1, \quad\forall t\in  [0,\pi ].
  $$
On the other hand, for any projection $E$ in $A$ satisfying
$$
  \text{rank}(E(t))=1, \quad\forall t\in  [0,\pi ],
  $$
the set
$$
  \mathcal E =\{(t, v)\in [0, \pi ]\times {\mathbb R}^2:E(t)v=v \}
  $$
is a one-dimensional, locally trivial vector bundle over $[0,\pi ]$.  Since all such vector bundles must be trivial, we are able to
find a continuous nonzero section  $e$ for $\mathcal E$.  This means that
$e(t)$ is a nonzero vector in the range of $E(t)$, for every $t$, which we may assume without loss of generality satisfies  $\|e(t)\|=1$, and
$e(0)=(1,0)$.
Since $E(\pi )=e_{11}$, we must have  $e(\pi )=\pm (1, 0)$, and it is possible to prove that the precise value of $e(\pi )$ is a
homotopy invariant of $E$.
Observing that this invariant  assigns $(1,0)$ to $P$, and $-(1,0)$ to $Q$, we see that $P$ and $Q$ are not homotopic to
each other.
