Problem with the proof of $g^{-1}(A\cap B)\subseteq g^{-1}(A)\cap g^{-1}(B)$.

Question

This is Abbott's exercise $1.2.7.b$. and it is not a duplicate of the following questions.

Proof of $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$

how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

My proof that $f^{-1}(D_1 \cap D_2) = f^{-1}(D_1) \cap f^{-1}(D_2)$

Exercise $1.2.7.b.$

The good behavior of preimages is completely general. Show that for an arbitrary function $g:R\rightarrow R$, it is always true that $g^{-1}(A\cap B)\subseteq g^{-1}(A)\cap g^{-1}(B)$.

I had no problem when proving it and I also checked the solution manual which its proof was similar to mine:

Let $x\in g^{-1}(A\cap B)$, then $g(x)\in(A\cap B)$. But this means...

My problem is with "Let $x\in g^{-1}(A\cap B)$, then $g(x)\in(A\cap B)$." What if our function isn't one-to-one? We chose our function arbitrarily so a simple function like $g(x)=x^2$ should be OK but it seems not to be.

Let's choose $A=[0,2]$ then $g(A)=[0,4]$ and $g^{-1}=[-2,2]$. So we can simply see that $x\in g^{-1}(A)$ does not imply $g(x)\in(A)$.

What is my mistake here?

Thank you in advance

Answer 1

For a function $f:R\rightarrow S$, and $D \subseteq S$, the inverse image $f^{-1}(D)$ is defined as follows: $$f^{-1}(D) = \{x \in R: f(x) \in D\}$$

By definition then, if $x \in g^{-1}(A\cup B)$ then $g(x) \in A \cup B$.

In your example, $f:\mathbb{R}\rightarrow \mathbb{R}$ given by $f(x) = x^2$, and $A = [0,2] \subseteq \mathbb{R}$.

If $x \in g^{-1}(A) = [ -\sqrt{2},\sqrt{2}]$. Then $g(x) \in [0,2]$, so there's no issue.

The problem seems to be that you're computing $g^{-1}(g(A))$, which is $g^{-1}([0,4]) = [-2,2] \neq A$. You are instead noticing that $g^{-1}(g(A)) \neq A$ (in general).

Answer 2

You definitely need to be a bit careful with these sets, because it's easy to accidentally make a false claim.

When you see $g^{-1}(A)$, its best to realize that there is only one thing it could be. Like, a priori, $g$ is a function from one set to another, so the symbol $g^{-1}$ doesn't automatically make sense. Nor does plugging a set into a function (rather than a point). So, by definition, when we say $g^{-1}(A)$, we mean the set of points which map into $A$ under $g$. That is, if $g\colon X\to Y$, $$g^{-1}(A)=\{x\in X:g(x)\in A\}.$$

Similarly, if $C\subseteq X$, then $$g(C)=\{y\in Y:y=g(c)\text{ for some }c\in C\}$$ or equivalently $$g(C)=\{g(c):c\in C\}.$$

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Now the claim is that $g^{-1}(A\cap B)=g^{-1}(A)\cap g^{-1}(B)$. To show set equality, we show that both inclusions hold.

First, take $x\in g^{-1}(A\cap B)$. Then, by definition, $g(x)\in A\cap B$, which means that $g(x)$ is in both $A$ and $B$. Now, by definition, $g(x)\in A$ means that $x\in g^{-1}(A)$, and $g(x)\in B$ means that $x\in g^{-1}(B)$. Thus $x\in g^{-1}(A)\cap g^{-1}(B)$.

Conversely, suppose $x\in g^{-1}(A)\cap g^{-1}(B)$. Then $x\in g^{-1}(A)$ means $g(x)\in A$, and $x\in g^{-1}(B)$ means $g(x)\in B$. Thus $g(x)\in A\cap B$, so by definition, $x\in g^{-1}(A\cap B)$.

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In your example, you take $g\colon (-\infty,\infty)\to [0,\infty)$ with $g(x)=x^2$. If $A=[0,2]$, then $g^{-1}(A)$ is the set of all points mapping into $[0,2]$. Thus, $g^{-1}(A)=[-\sqrt 2,\sqrt 2]$. If we pick and point $x\in[-2,2]$, then it is certainly true that $g(x)\in[0,2]$.