# Are these variations of replacement equivalent to the original, with and without separation and AC?

The axiom of replacement says, loosely speaking, something like "if a set surjects onto a class, that class is also a set." If we modify this as follows:

1. If a class injects into a set, that class is also a set.
2. If a class is in bijection with a set, that class is also a set.

then do we get the same thing as the original, given the other axioms of ZF(C)?

I'm particularly interested in how the answer changes depending on if we already have separation and are just changing replacement, vs if we only have replacement to begin with. Also interested in how things change with or without AC:

• If we have choice, I think the first modification is equivalent to the original, since every surjective function has an injective right inverse.
• If we have separation, I think the two modifications are the same as each other, but I'm not sure if they're equivalent to the original replacement.
• With both separation and choice I think all three forms of replacement are the same.

Without separation, the second modification seems to vary with things like GCH. This is because $$\Bbb N$$ exists, and thus every countable class in bijection with it must be a set. So we have all subsets of $$\Bbb N$$, and via the power set axiom, $$2^\Bbb N$$ exists. We've now managed to build $$\beth_0$$ and $$\beth_1$$, and any class in bijection with those must be a set. But if there's some subclass of $$\beth_1$$ of intermediate cardinality - is it required to be a set? It wouldn't be in bijection with either of those. This isn't a problem if we have the "injection" version, though, and it also isn't a problem if no class of intermediate cardinality exists.

Another way to put it: I'm not sure if we're able to build $$\aleph_1$$ with the "bijection" version if we don't already have separation (or possibly choice).

• Proper classes don't exist in ZF(C) and you seem to be talking about classes and proper classes here. So if a class injects into a set, does it mean there is a set describing the injection? Commented Aug 12 at 1:46

Separation makes the following three schemes equivalent:

1. If a set $$s$$ definably-surjects onto a class $$C$$, then $$C$$ is a set. (This is a scheme parameterized by the formulas giving the definable surjection and the class $$C$$. The other two versions are schematized similarly.)

2. If a set $$s$$ definably-bijects onto a class $$C$$, then $$C$$ is a set.

3. If a class $$C$$ definably-injects into a set $$s$$, then $$C$$ is a set.

For example, to get 1 from 2 we reason as follows. Suppose $$F$$ is a definable surjection of a set $$s$$ onto a class $$C$$. Let $$t$$ be the quotient of $$s$$ by the relation $$x\sim y\iff F(x)=F(y)$$; this exists via powerset and separation applied to $$s$$. Then $$F$$ yields a bijection from $$t$$ to $$C$$, so by 2 the class $$C$$ is a set. Similarly, to get 3 from 2, given a definable-injection $$I:C\rightarrow s$$ we can apply separation to $$s$$ to get the image of $$I$$, and then $$I$$ yields a definable bijection from $$C$$ to this set. Since the implications 3$$\rightarrow$$ 2 and 1$$\rightarrow$$2 are trivial, this completes the proof that all three are equivalent.

Note that choice wasn't needed here since we could use powerset to "quotient away" the bad functional behaviors. In $$\mathsf{ZFC}$$ without powerset, 2 and 3 are still equivalent but (I believe) 1 is strictly stronger.

• Do note that (3) implies Separation. If $\varphi(x)$ is a formula, define $\psi(x,p)$ to be $\varphi(x)\land x\in p$. Then if $A$ is a set, $\{x\mid\psi(x,A)\}$ is a class and it injects into $A$ for obvious reasons. Commented Aug 17 at 18:04