# $T_1 : V_1 \to V_2$ and $T_2 : V_2 \to V_1$ both onto. Then does it imply that $V_1$ and $V_2$ are isomorphic as vector spaces?

I was trying the following question :

Let $V_1$ and $V_2$ be two vector spaces such that there exists linear transformations $T_1 : V_1 \to V_2$ and $T_2 : V_2 \to V_1$ both onto. Then does it imply that $V_1$ and $V_2$ are isomorphic as vector spaces?

My attempt:

When $V_1$ and $V_2$ are finite dimensional then it's very easy to prove that the answer is affirmative.

But I am confused about the case where both are infinite dimensional. I really don't have any idea in this case.

Thanks in advance for help!

• You may be interested in Schroder-Bernstein theorem. If there exist injections $f:A\rightarrow B$ and $g:B\rightarrow A$ then there exists a bijection $h:A\rightarrow B$. – M. Nestor Apr 28 '18 at 15:55

## 1 Answer

Yes. In the infinite-dimensional case one can still show that any two bases have the same cardinality, and hence define the dimension to be the cardinality of a basis. Your hypothesis implies that $\dim(V_1)\ge\dim(V_2)$ and $\dim(V_2)\ge\dim(V_1)$, hence $\dim(V_1)=\dim(V_2)$, hence $V_1$ and $V_2$ are isomorphic (a bijection between the bases extends to an isomorphism).

• There are some subtleties on the difference between using surjections and using injections to compare cardinalities. Under the axiom of choice they are equivalent, but not without it. Does that affect your argument? – Arthur Apr 28 '18 at 16:03
• @Arthur I suppose. In my mathematical universe we assume AC without comment... – David C. Ullrich Apr 28 '18 at 16:22
• @Arthur. It has been shown to equi-consistent with ZF that (1) there exists a vector space without a Hamel (vector-space) basis, or (2) there exists a vector space with 2 bases that are cardinally not comparable. – DanielWainfleet Apr 28 '18 at 19:31
• "Well-order everything in sight"--- Mary Ellen Rudin. – DanielWainfleet Apr 28 '18 at 19:32