$f(A\cap B) = f(A)\cap f(B)$ if $f$ is surjective

Question

If $f$ is injective, then the statement $f(A\cap B) = f(A)\cap f(B)$ holds, but what if $f$ is surjective (and not injective)? Is the statement true as well?

I think that it is not true, but I am not sure whether my counterexample works: Assume $f: X\rightarrow Y$, where $X=[-2,2]$ and $Y=[0,4]$ and $f(x) = x^2$. Then $f$ is surjective. Now define $A=[-2,0]$ and $B=[0,2]$, then $f(A\cap B) = f(0) = 0$, whereas $f(A)\cap f(B) = [0,4]$.

Is this correct or is $f$ not surjective for the defined $A$ and $B$?

Your example is correct. It does not hold when $f$ is only surjective. — J1U, Oct 01 '18 at 16:06
Consider $X=A\cup B$ with $A\ne \emptyset \ne B$ and $A\cap B=\emptyset.$ And $Y={1}.$ And $f(A)=f(B)={1}.$ — DanielWainfleet, Oct 01 '18 at 22:28

score 1 · Answer 1 · answered Oct 01 '18 at 16:25

1

Your counter example works.

You may also consider the constant function $ f :\{ 1,2 \} \to \{3\}$ with $A=\{1 \}$ and $B=\{2\}$.

answered Oct 01 '18 at 16:25

Mohammad Riazi-Kermani

68,728

score 0 · Answer 2 · answered Oct 01 '18 at 16:25

When f is not injective we can not say the statement is true even if f is surjective as you see in your example.

The truth is $f(A\cap B)\subset f(A)\cap f(B)$. Let me prove: $y\in f(A\cap B)\iff \exists x\in A\cap B: y=f(x)\Rightarrow $ Since $x\in A$ and $ x\in B$ then $y=f(x)\in f(A)$ and $y=f(x)\in f( B) $$\Rightarrow y=f(x)\in f( A)\cap f(B)$

But we can't prove this $(f(A)\cap f(B)) \subset f(A\cap B)$. Here is the reason: Take $y\in f(A)\cap f(B)$ (we intend to say $y\in f(A\cap B)$). Then there is $a\in A$ and $b\in B$ such that $f(a)=f(b)=y$. Now to follow the proof we need to say $a=b$, but we can't. Here is the point why we need injectivity.

$f(A\cap B) = f(A)\cap f(B)$ if $f$ is surjective

2 Answers2