# Axiom of Choice and LEFT inverse [duplicate]

I am aware of why the Axiom of Choice is equivalent to the the statement that every surjection splits. However, I don't see why we don't also need AC to show that every injection splits.

In particular, for any injection $f$, we need to pick some element in the domain of $f$ to which elements in the codomain of $f$ are sent in the case that those elements are not in the image of $f$. This shows moreover that left inverses are not in general unique. How can we do this without AC?

EDIT: My real issue here is in the comparison of the proof of "$f: A \to B$ is a surjection implies $f$ has a right inverse" with the proof "$f: A \to B$ is an injection implies $f$ has a left inverse", and the use of the Axiom of Choice therein.

In particular, one apparently need not invoke AC in the injection proof to select in advance a single element $a_0 \in A$ for which to send $b \in B$ in the case that $b \notin f(A)$. But then, in the same way, in the surjection proof surely one may select for each $b \in B$ a single element $a_b \in f^{-1}[\{b\}]$ which then determines a function $g: B \to A$ such that $fg = \text{id}$, all without AC? We know each $f^{-1}[\{b\}]$ is not empty, so pick an element in $f^{-1}[\{b\}]$ (in the same way we did so in the injection proof by picking $a_0 \in A$) and send $b$ to that element.

## marked as duplicate by user228113, Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 20 '16 at 14:10

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• If $f:X\hookrightarrow Y$, you just pick any $x\in X$ and set $G|_{Y\setminus f(X)}$ constantly equal to $x$. You need to prove existence, not to find a canonical one. – user228113 May 20 '16 at 13:38
• If you had asked your question with this amount of details, I could close this as the proper duplicate. sigh – Asaf Karagila May 20 '16 at 15:38
• – Asaf Karagila May 20 '16 at 15:46

## 2 Answers

Say $f : X \to Y$, assume $X$ is nonempty. Pick any element $x_0$ in $X$. Since we are only choosing one element, we do not require the axiom of choice. Now set

$$g : Y \to X, ~~y \mapsto \begin{cases} x & \text{if}~~\exists x: f(x) = y \\ x_0 & \text{otherwise}.\end{cases}$$

This $g$ is a left inverse of $f$.

Pedantically, you say this:

Let $f\subseteq X\times Y$ an injective function $X\to Y$, where $X\neq\emptyset$.

Since $X\neq \emptyset$, there exists $x_0\in X$.

Let

$$G_{x_0}:=\left\{(y,x)\in Y\times X\,:\,(x,y)\in f\vee (x=x_0\wedge (\forall z\in X,\ (z,y)\notin f))\right\}$$

You can check that this is a function $Y\to X$ and, using the definition of composition, you can check that, for any $x_0\in X$, $G_{x_0}\circ f=id_{X}$.

As you can see, no AC used here, because (morally) you don't need it to pick exactly one element from a set.