4

This is a general question about homomorphisms on groups, rings, and fields.

If we are given a surjective homomorphism $f:A \rightarrow B$ and an injective homomorphism $g:A \rightarrow B$, can we say that A and B are isomorphic? Does the answer change depending on the structure of A and B, i.e. whether they are groups, rings, or fields?

Joe
  • 299

1 Answers1

6

This is false for groups and rings. For example, let $A=\prod_{i=1}^\infty \mathbb Z$ and $B=\mathbb Z/2\mathbb Z\times A$. Here, $A$ and $B$ can be viewed either as groups or as rings.

There is an obvious injection $A\to B$. There is also a surjection given by $$(a_1,a_2,\ldots)\mapsto (a_1\pmod 2, a_2,a_3,\ldots).$$ But certainly $A\not\cong B$, since the latter has an element of order $2$.

However, this is true for fields: if $g:A\to B$ is a surjective homomorphism of fields, then its kernel must be an ideal of $A$. Since $g$ is surjective, the kernel cannot be the whole of $A$. Hence $\ker g$ is trivial, so $g$ is also injective.

Mathmo123
  • 23,718
  • Notice also that in [the category of] topological spaces with continuous functions between them, a bijective continuous function need not be an isomorphism (for example, the identity from the discrete to the indiscrete topology on a space; its set-inverse is not continuous). – Patrick Stevens Apr 01 '16 at 17:31
  • @PatrickStevens that's true, but I'm not sure how that relates, since in all of these examples, a bijective morphism is an isomorphism. Equally, there are continuous injections and surjections from $\mathbb R\to\mathbb R^2$, but no continuous bijection (and certainly no homeomorphism). – Mathmo123 Apr 01 '16 at 17:40
  • 1
    I took the question to have general overtones of "how can morphisms fail to behave like set-maps?", but I take your point. – Patrick Stevens Apr 01 '16 at 17:42
  • 1
    This is true for vector spaces as well (by a dimension argument). – Crostul Apr 01 '16 at 18:01
  • 1
    For groups one can also make examples where they are both finitely generated and where one is finitely generated but the other is not (both the free group on $3$ and on countably many generators have injective and surjective maps to the free group on $2$ generators). – Tobias Kildetoft Apr 01 '16 at 19:12