Does inverse image of idempotent element contain an idempotent element? Suppose $f\colon \oplus A_i\to\oplus B_j$ is a ring homomorphism, where $A_i,B_j$ are local rings. Both sides have finitely many summands. Suppose an idempotent $x$ (i.e. $x=x^2$) is in the image of $f$. Can we always choose an idempotent in $f^{-1}(x)$?

Since $B_i $ are local, so the set of idempotents of $B_i $ has only 0 and 1. Thus $x=(x_1,x_2,...,x_n,...) $ where $x_i $are 0 or 1 and almost all of them are 0. So $x$ has an idempotents preimage. 
