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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?

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Perhaps you should clarify what is meant by "map" in this question. The answers using replacement seem to be assuming that the map is defined by a formula in the language of set theory. – user83827 Feb 23 '12 at 15:19
up vote 4 down vote accepted

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.

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The downvote seems a bit strange... – Asaf Karagila Aug 19 '13 at 0:37

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.

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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 $f$.

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Thanks! I'm glad you like it. – MJD Oct 15 '14 at 12:18

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