Is the domain of an one-to-one function a set if the target is a set?

Question

This is probably very naive but suppose I have an injective map from a class into a set, may I conclude that the domain of the map is a set as well?

Answer 1

If a function $f:A\to B$ is injective one, we can assume without loss of generality that $f$ is surjective too (by passing to a subclass of $B$), therefore $f^{-1}:B\to A$ is also a bijection.

If $B$ is a set then every subclass of $B$ is a set, so $f^{-1}:B\to A$ is a bijection from a set, and by the axiom of replacement $A$ is a set.

Answer 2

I say you can, since an injective map $f:A\to B$ is isomorphic to both its image (which is a set) and its domain.

Answer 3

Define $g$ on the range of $f$ such that $g(f(x)) = x$. This is well-defined because $f$ is injective. The domain of $g$ is equal to the range of $f$, which is a set.

Therefore by the axiom of replacement, or maybe the axiom of union applied to $\bigcup_{y\in{\rm Range}f} \{g(y)\} $, the range of $g$ is a set. But the range of $g$ is precisely the domain of $f$.