# Isomorphisms

Let $\phi : A_1 \rightarrow A_2$ be an F-Algebra homomorphism. I understand this to mean that $A_1$ and $A_2$ have the same algebraic structure (the meaning of which is also a little fuzzy to me…). Often we say in class “$\phi$ respects the operations.”

Is it true that the pre-image of $A_2$ under  $\phi$ contains (possibly multiple) pre-images of the basis vectors of $A_2$?

If so, then I have a follow  up:

We often try to show that $\phi$ is bijective in order to establish an isomorphism. Is this necessary? Would any bijective map $\psi$  from the basis of $A_1$ to the basis of $A_2$ extend linearly to a bijection between the algebras? Further, would the fact that they are homomorphic (by $\phi$) and the existence of some bijection $\psi$ between them establish an isomorphism even though $\psi$ and $\phi$ may be unrelated?

Lastly, must any homomorphism between isomorphic objects be bijective? (Does $\psi$ bijective imply $\phi$ bijective?)

1. To attempt to answer your last question: two objects $A, B$ are isomorphic if $\exists$ a bijection $\psi: A _1\to A_2$ such that $\psi(a)\psi(a')=\psi(aa')$, where $a, a'\in A_1$. So this means the map $\psi$ itself must be an isomorphism (a bijective homomorphism) by definition.
Thus, this seems to suggest that $\psi=\phi$. So not only must they be related, but are in fact identical.
• I meant “two objects $A_1, A_2$ are isomorphic…”
Let $A_1$ and $A_2$ be $F_2$, $F_2$-algebras over themselves and certainly isomorphic. Define $\psi(0)=1$ and $\psi(1)=0$. I think this map is bijective (though obviously not linear so not a homomorphism) and the image and pre-image are isomorphic, yet $\psi$ is not an isomorphism…