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…”

    • Briansdumb

      Is this a counter-example to your reply?

      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…

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