1
$\begingroup$

Generally speaking, a surjective function $f\colon A\to B$ has a right inverse requires AC to be valid. However, does proving that surjective linear transformation has a right inverse require Axiom of Choice? Since linear transformation is somewhat stronger than a general function. If so, how to make an elegant prove about that?

$\endgroup$
  • 1
    $\begingroup$ If $B$ is finite-dimensional, then you can choose a basis of $B$, and construct the right-inverse explcitly. If $B$ is infinite-dimensional, this approach requires AC to get the basis of $B$. $\endgroup$ – daw Sep 16 '17 at 10:28
  • 1
    $\begingroup$ math.stackexchange.com/questions/1713561/… is worth mentioning, as noted by @Jose in his answer. $\endgroup$ – Asaf Karagila Sep 16 '17 at 11:01
  • $\begingroup$ @AsafKaragila You're right. I'll add a remark about that. $\endgroup$ – José Carlos Santos Sep 16 '17 at 11:04
3
$\begingroup$

Yes, the axiom of choice is required.

Suppose that there is always such a right inverse $g\colon B\longrightarrow A$. In other words, there is always a linear map $g\colon B\longrightarrow A$ such that $f\circ g=\operatorname{Id}$. Let $C=g(B)$. Then $C\cap\ker f=\{0\}$. On the other hand, if $v\in A$, then $v=\left(v-g\bigl(f(v)\bigr)\right)+g\bigl(f(v)\bigr)$; but $v-g\bigl(f(v)\bigr)\in\ker f$ and $g\bigl(f(v)\bigr)\in C$. Therefore $C$ is a complement of $\ker f$. So, I proved that, if $f$ is a surjective linear map, then $\ker f$ has a complement.

But every vector subspace $V$ of $A$ is the kernel of some linear map; just take the natural projection from $A$ onto $A/V$. So, if the kernel of every linear map has a complement, then every vector subspace has a complement.

However, this assertion cannot be proved without the axiom of choice. This was proved more than a half-century ago by M. N. Bleicher, in “Some theorems on vector spaces and the axiom of choice” (Fund. Math. 54 (1964), 95–107).

Note: I suggest that you read this post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.