When an Isomorphism between two algebras is an equivalence?

Question

I'm currently involved in the study of algebraic structures, and there's a concept that seems to appear every so often. Given an Algebra $ A=<A,F_i> $ an isomorphism $ f $ is a bijective function that preserve the operations of the algebra. The example I want to recall to make my case involve direct representation of algebras. A direct representation of an algebra is an isomorphism between an algebra $ A $ and $ \prod(A,B_i) $ ($ B_i $ is a family of similar algebras to $ A $). It is known that if $ \theta_i=ker(f_i) $, then the image of the epimorphism $ f_i $ goes directly to $ A/\theta_i $ as for the first homomorphism theorem the conditions are satisfied and exists an immersion (that is an isomorphism, since they're all epimorphisms) from $ A/\theta_i $ and $ B_i $ of the direct representation. So at the end of the story, $ A $ is completely determined by its congruences, and $ A/\theta_i=B_i $ (not only isomorphic). The question rely exactly here: why this isomorphism is actually an equivalence? sumup of the question