Mathematics Asked by quanticbolt on December 14, 2020

Let the following be an axiom (which I will denote P):

If $x,y$ are sets and $f:yto x$ is a surjection, then the existence of an injection $f:xto y$ guarantees a choice function exists.

Does P lead to contradiction within ZF in any obvious Russell’s paradox like sense?

If I understand you correctly, you want to say that if there is a surjection from $y$ onto $x$, then if there is any injection, there is one splitting the surjection.

This cannot *lead* to contradiction, since it follows from the Axiom of Choice. Another question would be whether or not this implies the Axiom of Choice. This is a bit more complicated.

Let "The Partition Principle" denote the statement "if there is a surjection from $y$ to $x$, then there is an injection". And what we're asking is: Does the Partition Principle implies AC provable from $sf ZF$? And the answer is that we don't know.

Correct answer by Asaf Karagila on December 14, 2020

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