can i prove or reject it? Let $\Phi$ be a surjective map on an algebra $\mathcal{A}$ which satisfies the following condition for a fixed arbitrary $\epsilon \in \mathbb{C}$ which $\epsilon\neq1,-1$ and for fixed $P\in\{P_1,P_2\}$ such that $P_2=I-P_1$
$AP_i-\epsilon P_i A=\Phi(A)\Phi(P_i)-\epsilon \Phi(P_i)\Phi(A)$, $i\in\{1,2\} $
can i prove that $\Phi(I)\in \mathcal{Z}(\mathcal{A})$?
here I is unital element in $\mathcal{A}$, and $\mathcal{Z}(\mathcal{A})$ is the center of $\mathcal{A}$.
thanks for your help
 A: The key here will be establishing the conditions for $\epsilon =1$ (which is not assumed). It is not clear $\mathcal{A}$ is an algebra over which ring. I assume that it is over $\mathbb{C}$. 
To do this, we write for some $\epsilon\neq \pm1, 0$ with 
$$
AP_i - \epsilon P_i A = \Phi(A)\Phi(P_i) - \epsilon \Phi(P_i) \Phi(A)$$ 
and
$$
AP_i - \epsilon^{-1} P_i A = \Phi(A)\Phi(P_i) - \epsilon^{-1} \Phi(P_i) \Phi(A)$$
Then multiply $\epsilon$ to the second to obtain 
$$
\epsilon AP_i -   P_i A = \epsilon\Phi(A)\Phi(P_i) -   \Phi(P_i) \Phi(A)$$
Then we have 
$$(1+\epsilon)( AP_i - P_i A) = (1+\epsilon) (\Phi(A)\Phi(P_i)-\Phi(P_i)\Phi(A))$$
Since $\epsilon \neq -1$, we multiply $(1+\epsilon)^{-1}$ to obtain
$$
AP_i - P_iA = \Phi(A)\Phi(P_i)-\Phi(P_i)\Phi(A).$$
Then adding up the equations for $i=1, 2$, we have
$$
AI-IA = \Phi(A)\Phi(I) - \Phi(I)\Phi(A).$$
Since $I$ is the unity, $AI-IA=O$. Thus, we see that 
$$\Phi(A)\Phi(I) - \Phi(I)\Phi(A)=O.$$
Since $\Phi$ is surjective, we obtain that $\Phi(I)\in\mathcal{Z}(\mathcal{A})$. 
