Talk:Right Inverse Mapping is Injection

The theorem statement appears to read: "If $f: S \to T$ is a surjection, and if $g: T \to S$ is a right inverse of $f$, then $g$ is an injection." So why does this theorem depend on the existence of such a $g$? --abcxyz (talk) 15:36, 12 October 2012 (UTC)


 * It doesn't; the page needs to be rewritten (we have come to avoid theorem statements like "Any (thing) satisfying (premises)" and rather replace them with "Let (thing) satisfy (premises)"); in particular, the reference to Surjection iff Right Inverse needs to be removed. Maybe even the condition that $f$ be surjective can be removed (this follows from the existence of a right inverse by the half of Surjection iff Right Inverse that doesn't depend on AC). Good call. --Lord_Farin (talk) 16:09, 12 October 2012 (UTC)


 * Yes I understand where you're coming from now. No worries. --prime mover (talk) 18:47, 12 October 2012 (UTC)


 * Still, I'd say the condition that $f$ be a surjection should be dropped; it isn't even used (of course, it's implied, but sometimes one doesn't want to derive separately that $f$ is a surjection when the right inverse $g$ obviously works. --Lord_Farin (talk) 20:54, 12 October 2012 (UTC)


 * Sorted. This now matches Left Inverse Mapping is Surjection. --Dfeuer (talk) 17:54, 1 May 2013 (UTC)