Indirectly inspired by this post we ask the following question:
Let $A$ be a $C^*$ algebra which is equiped with a faithful positive normal trace. Assume that $e_1,e_2,f_1,f_2$ are idempotents in $A$. Assume that $e_1 \simeq_h f_1$ and $e_2\simeq_h f_2$ where $\simeq_h$ is the homotopy equivalent of idempotents. Does this imply that $\operatorname{trace}(e_1e_2)=\operatorname{trace}(f_1f_2)$?
I have never seen the notion of homotopy of idempotents, but I will assume what I think it should be: $e\simeq_h f$ means there exists $F:[0,1]\to A$, continuous, such that $F(0)=e$, $F(1)=f$.
In that case, the answer is no. Take $A=M_2(\mathbb C)$, and $$ e_1=f_1=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\quad e_2=\begin{bmatrix} t&\sqrt{t-t^2}\\ \sqrt{t-t^t2}&1-t\end{bmatrix} ,\quad f_2=\begin{bmatrix} s&\sqrt{s-s^2}\\ \sqrt{s-s^t2}&1-s\end{bmatrix}, $$ where $t,s\in[0,1]$ are to be chosen. Then all four projections are homotopic, and $$ \operatorname{Tr}(e_1e_2)=t,\qquad \operatorname{Tr}(f_1f_2)=s. $$