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I asked a question Under what condition does $f(f^{-1}(f(A))) = f(A)$? and it totally backfired because people were confused whether $f^{-1}$ is the preimage or the inverse function

Let $f: X \to Y$ be a function, $A\subseteq X$

So what is the difference between:

$(f^{-1})^{-1}(A)$ and $f(f^{-1}(f(A)))$?

And in general what property does $f$ have to satisfy in order $(f^{-1})^{-1}(A) = f(A)$?

Thanks!

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Olórin
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  • You'd have to define $\left(f^{-1}\right)^{-1}$. $g^{-1}$ is generally defined on subsets of the range of a function, $g$, so it is unclear what is meant by $\left(f^{-1}\right)^{-1}(A)$. – Thomas Andrews Jun 09 '16 at 22:44
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    What do you understand $(f^{-1})^{-1}(A)$ to mean when $f(x)$ is not a bijection? – Henry Jun 09 '16 at 22:46
  • @Henry I was hoping to write the $(f^{-1})^{-1}$ as a composition like $f(f^{-1}(f))$ so I can see what happens if $f$ is bij, inj, or surj – Olórin Jun 09 '16 at 22:48
  • Do you intend for $f^{-1}(A)$ to mean the pre-image of $A$ or the image of $A$ under the map $f^{-1}$? – T.J. Gaffney Jun 09 '16 at 22:49
  • @Gaffney The problem is that I didn't write it, I was just given a question when $(f^{-1})^{-1}(A) = f(A)$ – Olórin Jun 09 '16 at 22:50

2 Answers2

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The trouble is in ambiguity. Notation $\cdot ^{-1}$ can mean both inverse and preimage, when it comes to functions. Also note that if $f\colon X\to Y$, $f^{-1}$ can denote a function $f^{-1} \colon \mathcal P(Y)\to\mathcal P(X)$ that assigns to each subset of $Y$ its preimage under $f$. Thus, $(f^{-1})^{-1}$ could be four different things:

1) Inverse of inverse:

In this case $(f^{-1})^{-1} = f$ by uniqueness of inverse. Thus, $(f^{-1})^{-1}(A) = f(A)$.

2) Preimage of inverse:

In this case, since $f$ is invertible, the preimage is just the image of $f$. Thus, $(f^{-1})^{-1}(A) = f(A)$.

3) Inverse of the above preimage function.

We will inspect this shortly.

4) Preimage of the above preimage function:

This, I think, would just be strange, unless this comes from some set theory course, and in this case $(f^{-1})^{-1}(A) = f(A)$ wouldn't even make sense since $A$ would need to be subset of $\mathcal P(X)$ and not subset of $X$ which is required for $f(A)$ to make sense.

The only interesting case is 3), and let us prove two things:

a) $f^{-1}$ is injective $\implies$ $f$ is surjective

b) $f^{-1}$ is surjective $\implies$ $f$ is injective

Assume the $f$ is not surjective. Then, $f^{-1}(f(X)) = X = f^{-1}(Y)$, but $f(X)\neq Y$, so $f^{-1}$ is not injective. This proves a) by contraposition.

Now, assume that $f^{-1}$ is surjective and let $x_1, x_2\in X$ such that $f(x_1) = f(x_2)$. Since $f^{-1}$ is surjective, there is an $A\subseteq Y$ such that $f^{-1}(A) = \{x_1\}$. Obviously, $f(x_2) = f(x_1)\in A$, by definition, but then $x_2\in f^{-1}(A) = \{ x_1\}$, and thus $x_1 = x_2$, so $f$ is injective.

Combining a) and b), we have that if the preimage $f^{-1}$ is bijective, so is $f$. We then claim that the inverse of the preimage $f^{-1}$ is just the image of $f$. But this is obvious since for invertible $f$ we have that $f(f^{-1}(A)) = A$ and $f^{-1}(f(A)) = A$.

Thus, in case 3), again we have $(f^{-1})^{-1}(A) = f(A)$.

In conclusion, $(f^{-1})^{-1}(A) = f(A)$ makes sense if and only if $f$ is invertible, and in all meaningful interpretations is true.

Ennar
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You are confused by the two different meanings of $f^{-1}(B)$ when $B\subset Y$. But actually this ambiguity does not exist, because if $f$ is a bijection, so that the inverse function $f^{-1}:Y\to X$ exists, then the image of $B$ under $f^{-1}$ is the same thing as the preimage under $f$. This is in fact the point 2) in @Ennar's answer, but I thought it might help you to see this part in more details.

Indeed, if $x$ is in the preimage of $B$ under $f$, then $y=f(x)\in B$ and thus $x=f^{-1}(f(x))=f^{-1}(y)$ is in the image of $B$ under $f^{-1}$.

Moreover, if $x=f^{-1}(y)$ is in the image of $B$ under $f^{-1}$, then $f(x)=f(f^{-1}(y))=y\in B$, so $x$ is in the preimage of $B$ under $f$.

Maybe I should have mentionned this in my answer to your original question. Sorry about the confusion.

Arnaud D.
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