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A (very?) naΓ―ve question, but I didn't get an answer on math.se: so here goes ….

In his original ETCS paper, Lawvere states a categorial version of choice for sets in this form: "If the domain of $f$ has elements, then there exists $g$ such that $fgf = f$". So let's say, generalizing

(1) A category has Choice-1 iff for any $f\colon X \to Y$ where $X \not\cong 0$ [with $0$ initial], there exists a $g \colon Y \to X$ such that $f \circ g \circ f = f$.

Another more familiar version of choice for sets is that any surjection has a right inverse (or left inverse, depending on your preferred handedness convention!). Generalizing, let's say

(2) A category has Choice-2 iff for any epic $f\colon X \to Y$ there exists a $g \colon Y \to X$ such that $f \circ g = 1_Y$.

But (forgive a senior moment!) I'm embarrassingly unclear about the conditions β€” the weakest/most natural/nicest conditions? β€” under which a category which has Choice-2 has Lawvere's Choice-1. Is there a standard story about this which I've missed?

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    $\begingroup$ I don't think that (1) implies (2) in any category, because for an epimorphism $0 \to Y$ the condition (1) gives no information. So, for example, for the category $\bullet \to \bullet$ condition (1) is trivially satisfied, but (2) is not. $\endgroup$
    – Arshak Aivazian
    Commented Jun 27, 2023 at 7:33
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    $\begingroup$ OOOPS! Initially I wrote that if a category has Choice-1, it trivially has Choice-2, which was just silly as @Arshak points out, as we only get the trivial implication for the cases where 𝑋≇0! But the remaining case is OK in a topos, say: for then any arrow 0→𝑋 is monic, so an epic 𝑓:0→𝑋 will be iso, since a topos is balanced, and so $f$ will have the required inverse $\endgroup$
    – Peter Smith
    Commented Jun 27, 2023 at 13:28
  • $\begingroup$ There is also a parametrized version: for any epi $X\to Y$ and any $U$ there is an epi $V\to U$ and a morphism $V\times Y\to X$ with $V\times Y\to X\to Y$ equal to the product projection. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 2, 2023 at 5:58

1 Answer 1

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If we take Choice-1 in Lawvere's original form with "has (global) elements" in place of "is not initial", then Choice-2 follows from Choice-1 in any Boolean category. Factor $f:X\to Y$ as an epi $h : X \twoheadrightarrow Z$ followed by a mono $m:Z\hookrightarrow Y$. Then $h$ is a split epi by Choice-2, while $m$ is complemented by Booleanness, and therefore a split mono since $Z$ has a global element. So we have $s:Z\to X$ with $h s = 1_Z$, and $r:Y\to Z$ with $r m = 1_Z$, and defining $g = s r$ we have $f g f = (m h) (s r) (m h) = m (h s) (r m) h = m h = f$.

In a topos, Booleanness also follows from Choice-1 by the Diaconescu-Goodman-Myhill theorem. So Choice-2 also follows from Lawvere's original Choice-1 in any topos.

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  • $\begingroup$ Ok, thanks! Let's clean up our comments, now :-) $\endgroup$
    – David Roberts ♦
    Commented Jul 3, 2023 at 2:02

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