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Let $f : A \rightarrow B$, and let $C \subseteq A$ and $D \subseteq B$. Prove:

  1. $f^{-1}(f(C)) \supseteq C$
  2. $f(f^{-1}(D)) \subseteq D$

My instructor gave the following definitions for image and pre-image:

  • The image of C under $f$ is the set $f(C) = \{f(x) : x \in C \}$
  • The pre-image of D under $f$ is the set $f^{-1}(D) = \{x : f(x) \in D \}$

Proof of (1):

Let $x \in C \implies f(x) \in f(C) \implies x \in f^{-1}(f(C))$. Therefore $C \subseteq f^{-1}(f(C))$ and equivalently, $f^{-1}(f(C)) \supseteq C$.

Proof of (2):

Let $f(x) \in f(f^{-1}(D)) \implies x \in f^{-1}(D) \implies f(x) \in D$. Therefore $f(f^{-1}(D)) \subseteq D$.

My issue that I'm having is understanding why the definition of image and pre-image do not let us conclude that $f^{-1}(f(C)) = C$ and $f(f^{-1}(D)) = D$. I came up with counter-examples for both of these and I understand that they are only true when $f$ is injective and surjective, respectively. However in the the above proofs I cannot see why the arrows do not go the other way. In particular it would seem that in (1) we cannot say that $f(x) \in C \implies x \in C$ and in (2) we cannot say that $f(x) \in D \implies x \in f^{-1}(D)$. Why do the definitions given above not permit these implications?

Dirk Mahogany
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  • https://math.stackexchange.com/a/359694/334795 has a solution to your questions. – Hermis14 Dec 22 '21 at 00:30
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    $f(x)\in f(C)$ only implies that there exists $c\in C$ such that $f(x)=f(c)$. You cannot conclude from this that $x\in C$, unless $f$ is one-to-one, since then you can go from $f(x)=f(c)$ to $x=c$ to $x\in C$. That's why $f(x)\in f(C)$ does not imply $x\in C$. Now, you can say that if $f(x)\in D\iff x\in f^{-1}(D)$, by definition of $f^{-1}$. What you cannot do is say that if $d\in D$ then $d=f(x)$ for some $x\in f^{-1}(D)$. So you can't bridge the gap between $f(f^{-1}(D))\subseteq D$ to $D\subseteq f(f^{-1}(D))$. – Arturo Magidin Dec 22 '21 at 00:37
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    In other words, the second part you do have a biconditional: $f(x)\in D\iff x\in f^{-1}(D)$. But this only shows that $f(f^{-1}(D)) = D\cap\mathrm{Im}(f)$, it does not prove that $D=f(f^{-1}(D))$, because you do not know that every element of $D$ is of the form $f(x)$ for some $x\in X$ unless $f$ is surjective. – Arturo Magidin Dec 22 '21 at 00:39

2 Answers2

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If $f(x) \in C$, it is indeed true that it does not follow that $x \in C$. Think about the function $\mathbb{R} \ni x \mapsto x^2\in \mathbb{R}$. Even though $f(\sqrt{2}) = 2 \in \mathbb{Q}$, obviously $\sqrt{2} \notin \mathbb{Q}$. Why should the image of an element being in a set mean that this element itself is in the set? Functions can be very weird and send sets to all kinds of other sets. This is where the reciprocal implication fails in $(1)$.

On the other hand, $f(x) \in D$ does indeed imply that $x \in f^{-1}(D)$. This is because we define $f^{-1}(D)$ to be the set of all elements whose images are in $D$, that is: $x \in f^{-1}(D) \iff f(x) \in D$. So that's clearly not the problem here. It's not that $x \in f^{-1}(D) \implies f(x) \in f(f^{-1}(D))$ is false either, that's clearly true (because $a \in A$ clearly implies $f(a) \in f(A)$ always). All this can only mean one thing: your proof for $(2)$ is wrong. Because if it was right, we would get the reciprocal implication for free. The problem here is the same one as in $(1)$: $f(a) \in f(B)$ does not imply $a \in B$.

For the correct proofs and more details, see here.

Matheus Andrade
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For prove $f^{-1}(f(C))\subseteq C$, you need prove that for all $x\in f^{-1}(f(C))$ should belong to an $C$, right?

Then $$x\in f^{-1}(f(C))\Rightarrow f(x)\in f(C)\Rightarrow \exists y\in C \text{ s.t. }f(x)=f(y)$$ At this point I can not conclude since who belongs to the set $C$ is $y$, but I do not know anything about $x$. But things change if $f$ is injective, since in this case $f(x)=f(y)\Rightarrow x=y\in C$, and therefore $x\in C$. Taking this into account yourself, you can give a counterexample that you need, I think that understanding the need for injectivity is much more productive.

Now, for prove $D\subseteq f(f^{-1}(D))$, In much the same way, we start from $y\in D$, but this element may or may not be an image of some element of $C$, that is, it may happen that $y\in D\setminus f(C)$. So we cannot say anything, but everything changes if we assume that $f$ is surjective since in this case we have $\exists x\in C$ such that $y=f(x)$, then $x\in f^{-1}(D)$. Therefore $y=f(x)=f(f^{-1}(D))$. Again the need for the surjectivity of $f$ is evidenced.

Zaragosa
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