My question is whether the axiom of extensionality is required to show that the schema of collection follows from the schema of replacement in the usual ZermeloFraenkel environment with choice. In other words: Is the schema of collection a theorem schema in ZermeloFraenkel set theory with choice minus the axiom schema of extensionality?

1$\begingroup$ Why would you remove extensionality from ZFC? $\endgroup$ – François G. Dorais♦ Dec 20 '12 at 2:56

$\begingroup$ Extensionality is not a schema. (Well, I suppose it is a trivial schema with only one instance.) $\endgroup$ – François G. Dorais♦ Dec 20 '12 at 2:57

2$\begingroup$ At first when I read this I thought "what the heck?" but then I realized it is kind of an amusing question. Without extensionality, there's not really any way to prove that two sets are equal, and hence not much of any way to construct anything unique. So replacement appears to become kind of useless. (: $\endgroup$ – Mike Shulman Dec 20 '12 at 5:30

2$\begingroup$ Hello Frode, welcome to MO! $\endgroup$ – Emil Jeřábek supports Monica Dec 20 '12 at 12:33

2$\begingroup$ A related fact would be that one does need the power set axiom to prove collection from replacement. See jdh.hamkins.org/whatisthetheoryzfcwithoutpowerset. $\endgroup$ – Joel David Hamkins Dec 20 '12 at 13:48
Collection is not provable in ZFC minus extensionality, a simple countermodel is described in https://mathoverflow.net/questions/54328 . (That the model cannot provably satisfy collection follows from Gödel’s theorem. For a specific instance of collection which fails, let $\bar\omega$ denote one of the many representations of $\omega$ in the model, and $\bar0\in\bar\omega$ the corresponding empty set: then the model satisfies “for every $n\in\bar\omega\smallsetminus\{\bar0\}$, there exists a function $f$ with domain $n$ such that $f(\bar0)=\bar\omega$, and $f(x)\in f(y)$ whenever $x\in y\in n$”, but there is no set collecting such functions for every $n\in\bar\omega\smallsetminus\{\bar0\}$.)