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A (very?) naive 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 -- the weakest/most natural/nicest conditions? -- under which a category which has Choice-2 has Lawvere's Choice-1. Is there a standard story about this which I've missed?

A (very?) naïve 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 — the weakest/most natural/nicest conditions? — under which a category which has Choice-2 has Lawvere's Choice-1. Is there a standard story about this which I've missed?

A (very?) naive 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 $f \circ g = 1_Y$.

Now, trivially, if a category has Choice-1, it has Choice-2 [added, at least for $X \not\cong 0$]. But (forgive a senior moment!) I'm embarrassingly unclear about the conditions -- the weakest/most natural/nicest conditions? -- under which a category which has Choice-2 has Choice-1. Is there a standard story about this which I've missed?

A (very?) naive 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 -- the weakest/most natural/nicest conditions? -- under which a category which has Choice-2 has Lawvere's Choice-1. Is there a standard story about this which I've missed?

A (very?) naive 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 $f \circ g = 1_Y$.

Now, trivially, if a category has Choice-1, it has Choice-2. But (forgive a senior moment!) I'm embarrassingly unclear about the conditions -- the weakest/most natural/nicest conditions? -- under which a category which has Choice-2 has Choice-1. Is there a standard story about this which I've missed?

A (very?) naive 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 $f \circ g = 1_Y$.

Now, trivially, if a category has Choice-1, it has Choice-2 [added, at least for $X \not\cong 0$]. But (forgive a senior moment!) I'm embarrassingly unclear about the conditions -- the weakest/most natural/nicest conditions? -- under which a category which has Choice-2 has Choice-1. Is there a standard story about this which I've missed?