# How do you indicate that you're asking for a left or a right inverse in a question?

I've been asked the question:

For a function $$f$$ from X $$\to$$ Y, $$f^{-1}(f(A)) = A$$ for every subset A of $$X$$ if and only if: (a) $$f$$ is injective (b) $$f$$ is surjective (c) $$f$$ is bijective.

A function is invertible only if it's bijective, and so I thought that the answer would be (c). But the correct answer, apparently, is (a).

Is there anything in the question that makes it clear that what's being asked is a left inverse, in which case it's sufficient for the function to be only injective? I'm guessing it has something to do with the statement "for every subset A of X" -- am I correct, or is there something else?

Also, if you were to specify "right inverse", how would you modify the question?

• Why did you write "apparently the answer is (a)"? if $f(x)=y\in f(A)$ such that $x\in A$, shouldn't that be enough for $f$ being bijective? The only exception would be the restriction of 1-1: Exists an $y\in f(A)$ such that $f^{-1}(y) \ne x \in A$. – Lincon Ribeiro Feb 23 at 17:37
• @LinconRibeiro See my answer - you cannot in fact conclude bijectivity from the OP's property. – Noah Schweber Feb 23 at 17:56

## 1 Answer

Remember that the notation "$$f^{-1}(S)$$" refers to the set $$\{x\in dom(f): f(x)\in S\}.$$ No assumptions on $$f$$ need to be made for this to make sense. In particular, using this notation does not presuppose that $$f$$ has an inverse.

Similarly, "$$f(S)$$" refers to the set $$\{x\in codom(f): \exists s\in S(f(s)=x)\}$$ (or, perhaps more simply, $$\{f(s): s\in S\}$$).

Unwinding definitions a bit, the problem here is just asking about the property

$$(*)\quad$$ For every set $$A\subseteq dom(f)$$ we have $$\{x\in dom(f): f(x)=f(a)\mbox{ for some a\in A}\}=A.$$

Now let's see why the answer to the question is (a) - that property $$(*)$$ holds iff $$f$$ is injective.

First, let's show that injectivity of $$f$$ implies $$(*)$$. It's helpful to note that we trivially have $$A\subseteq \{x\in dom(f): f(x)=f(a)\mbox{ for some a\in A}\}$$, so to show that property $$(*)$$ holds we only need to pay attention to the reverse inclusion. Suppose $$f$$ is injective and let $$A\subseteq dom(f)$$ be arbitrary. Let $$x\in dom(f)$$ such that $$f(x)=f(a)$$ for some $$a\in A$$. Since $$f$$ is injective, $$x$$ must actually be this $$a$$; that is, if $$f(x)=f(a)$$ for some $$a\in A$$ then $$x\in A$$ already. So the left hand side set is a subset of $$A$$, and per the comment above we're done.

Note that this is the surprising direction - that injectivity alone guarantees a property which at first glance may seem to be bijectivity. So before going on to the next point, make sure the argument above makes sense.

Now let's show that $$(*)$$ implies injectivity. Suppose for every $$A\subseteq dom(f)$$ we have $$\{x\in dom(f): f(x)=f(a)\mbox{ for some a\in A}\}=A$$. Fix $$u,v\in dom(f)$$ such that $$f(u)=f(v)$$; we want to show $$u=v$$. Let $$A=\{u\}$$. Then $$\{x\in dom(f): f(x)=f(a)\mbox{ for some a\in A}\}$$ contains $$v$$, so by our assumption $$v\in A$$; but $$A=\{u\}$$, so $$v=u$$.