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This might be a stupid question. But I am reading a math book and we have to show that: $$f^{-1}(C\cap D)=f^{-1}(C) \cap f^{-1}(D)$$ I have no problem proving that and that question has already been asked. But then we are supposed to show that $$f(C\cap D)\neq f(C) \cap f(D)$$ by counterexample. I found answers on that question as well.

But this is kind of weird. Shouldn't the first identity also hold for $f$? Because after all $f$ and $f^{-1}$ are just functions.

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    $\begingroup$ $f^{-1}$ is not “just a function”. It’s the complete inverse image, which behaves quite a bit nicer than functions. $\endgroup$
    – Milten
    Sep 13, 2021 at 17:11
  • $\begingroup$ To be sure, if $f:X\to Y$, then $f^{-1}$ is a function, but it’s not $Y\to X$, but rather $P(Y)\to P(X)$. $\endgroup$
    – Milten
    Sep 13, 2021 at 17:14
  • $\begingroup$ @AdamRubinson math.stackexchange.com/questions/482633/… $\endgroup$
    – eeqesri
    Sep 13, 2021 at 18:37
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    $\begingroup$ Because $f^{-1}$ must be injective: else $f$ wouldn’t be a function. $\endgroup$ Sep 13, 2021 at 18:42
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    $\begingroup$ @AdamRubinson $f^{-1}$ is only injective if $f$ is surjective. $\endgroup$
    – Rob Arthan
    Sep 13, 2021 at 19:59

2 Answers 2

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Disclaimer: As a learner, I found this question confusing as well and was hardly satisfied with answers I could find. Then I posted an answer here myself, unfortunately that was wrong. But after deleting it and studying this again, I now have a better, even if lengthy, answer aimed at confused learners. Please let me know if there are still problems.

Clarification of question, intuition and sources

To clarify, in my book, it is formulated as:

Let $A, B$ be sets and $M_1$, $M_2$ $ \subseteq A$, as well as $N_1,N_2$ $ \subseteq B$. Let $f: A \mapsto B$.

Then it is asked, among other things,:

(1) Show $f^{-1}(N_1 \cap N_2) = f^{-1}(N_1)\cap f^{-1}(N_2)$

(2) Does in general also $f(M_1 \cap M_2) = f(M_1)\cap f(M_2)$ hold ?

Now, why does it hold for the inverse in (1) of $f$ but not $f$ itself in (2)? In (2) in this direction $\implies$ the domain is transformed through the map and you select an element $f(x)$ from $f(M_1 \cap M_2)$ so $x$ must be $ \in M_1 \cap M_2$ but the image of those two sets, when taken separately, might behave differently than the domain subsets. Hence you do not "know" yet what the map does to the elements and this is the exact problem you do not have with the inverse image where both intersections are predefined so to speak. I think it is easiest to understand through a couple of examples and visually. But first here are a few mathstackexchange links that clarify the question rigorously: A nicely written proof of (1). A proof of why (2) holds if and only if $f$ is injective.

First a counterexample with respect to (2)

We can easily construct a counterexample:

Let $A = \{1,2\}, B = \{3\}, M_1 = \{1\}, M_2 = \{2\}, f(1) = 3 = f(2)$. Now here $f(M_1 \cap M_2) = \emptyset $ and $f(M_1) \cap f(M_2) = \{3\}$.

Note that this function specifically is not injective. And also note that still $f(M_1 \cap M_2) \subseteq f(M_1)\cap f(M_2)$ (here because $\emptyset$ is a subset of any set) but $f(M_1)\cap f(M_2)\nsubseteq f(M_1 \cap M_2)$.

Clarifying examples

Looking at the example we used as a counterexample above, we can observe that (1) still holds here because $N_1 = N_2 = \{3\}$ and so $f^{-1}(N_1 \cap N_2) = \{1,2\}$. Also $f^{-1}(N_1) = \{1,2\} = f^{-1}(N_2)$.

Let us take a look at another example. Let $A=\{1,2,3,4\}$ and $B=\{a,b,c,d\}$ as well as $M_1 = \{1,2\}$, $M_2 = \{3,4\} \subseteq A$, as well as $N_1 = \{a,b\},N_2 = \{c,d\} \subseteq B$. Define $f: A \mapsto B$ via $f = \{(1,a),(2,d),(3,d),(4,d)\}$.

So all elements in $A$ are mapped to $d$ except $1$ to $a$. This function is clearly neither injective nor surjective. But (1) still holds. (2) is clearly violated again because $f(M_1 \cap M_2) = \emptyset \neq f(M_1) \cap f(M_2) = \{d\}$

If however you had an injective function, (2) would hold. For instance adjust the counterexample so that you have an injective function by adding an element to the codomain and changing the map somewhat:

Let $A = \{1,2\}, B = \{3,4\}, M_1 = \{1\}, M_2 = \{2\}, f(1) = 3, f(2) = 4$. Now here $f(M_1 \cap M_2) = \emptyset $ and $f(M_1) \cap f(M_2) = \ \emptyset$. So they are equal.

Schematics related to (2) and then (1)

First visualize a non-injective map.

enter image description here

Make sure you understand that if $f$ is injective (2) does hold.

enter image description here

In the inverse image if you select an element first from $f^{-1}(N_1 \cap N_2)$ or from $f^{-1}(N_1)\cap f^{-1}(N_2)$ does not matter since there is a direct correspondence or equivalence between these intersections (which is what you prove anyway of course). So in the end it comes down I think to really understanding the notation of the inverse image.

enter image description here

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  • $\begingroup$ this is a really beautiful answer. I particularly like the diagrams. Thanks. $\endgroup$
    – eeqesri
    Oct 18, 2023 at 15:34
  • $\begingroup$ Hold it initially I thought everything was right. But I noticed something I didn't understand completely. You wrote in the first example of the clarifying example that $f^{-1}(N_1)=f^{-1}(N_2)=\{1,2\}$ but how can that be when $N_1=N_2=\{3\}$. I think one can't have distinct values for $f^{-1}(3)=\{1,2\}$ $\endgroup$
    – eeqesri
    Oct 18, 2023 at 15:47
  • $\begingroup$ I hope it is helpful. Yes you can, see here the definition of the inverse image of a set $B$ with respect to $f^{-1}(B)$. I denoted this here for the sets $N_1$ and $N_2$. en.wikipedia.org/wiki/Image_(mathematics)#Inverse_image The inverse image of the set $\{4\}$ for instance is $\{-2,2\}$ with respect to the function $f(x)=x^2, x \in \mathbb{R}$. $\endgroup$
    – yomath
    Oct 18, 2023 at 18:49
  • $\begingroup$ It is also like the first diagram, because $f$ is not injective after all, so several values in the domain relate to one specific codomain value. $\endgroup$
    – yomath
    Oct 18, 2023 at 18:57
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Let us consider the two statements

$$f^{-1}(C\cap D)=f^{-1}(C) \cap f^{-1}(D) \tag{1}$$ and $$f(A\cap B)=f(A) \cap f(B) \tag{2} .$$ You know that $(1)$ is true for all $C, D \subset Y$, but $(2)$ is in general not true for all $A, B \subset X$.

One can easily prove that $(2)$ is true for all $A, B \subset X$ if and only if $f$ is injective, but that is another question. You ask

But this is kind of weird. Shouldn't the first identity also hold for $f$? Because after all $f$ and $f^{-1}$ are just functions.

Your question is a bit sloppy, so let me reformulate it.

Let $\mathfrak P(R)$ denote the power set of a set $R$.

Let us consider a function $$\phi : \mathfrak P(R) \to \mathfrak P(S)$$ between two power sets. We say that $\phi$ preserves intersections if $$\phi(U \cap V) = \phi(U) \cap \phi(V) \text{ for all } U, V \in \mathfrak P(R) .\tag{3}$$ Now each function $f : X \to Y$ between sets $X, Y$ induces two functions between power sets $$f^{-1} : \mathfrak P(Y) \to \mathfrak P(X), f^{-1}(C) = \{ x \in X \mid f(x) \in C \},$$ $$f : \mathfrak P(X) \to \mathfrak P(Y), f(A) = \{ f(x) \mid x \in A \}.$$

What you ask is

Why does $f^{-1}$ preserve intersections, but $f$ does not? Because after all $f$ and $f^{−1}$ are just functions.

Yes, both $f^{-1}$ and $f$ are functions between power sets. But there is no reason why a given function $\phi$ between power sets should preserve intersections. This is a very special property of $\phi$. And the simple truth is that $f^{-1}$ has it, but $f$ does not.

By the way

  • $f^{-1}$ preserves intersections of arbitrary families of subsets of $Y$
  • Both $f^{-1}$ and $f$ preserve unions of arbitrary families of subsets of $Y$ and $X$, respectively.
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