Image of the intersection is contained in the intersection of images; why are not these equal?

Question

If there are any minor mistakes in my proof, it would be great if they were pointed out - but let it not be the central discussion. I'm rather concerned why the answer is $\subset$ instead of $=$ since I have proved it both ways. It's not immediately clear to me.

Prove $f(\cap \scr{C}) \subset \cap f(\scr{C})$.

Here are the conditions:

Let $y \in f(\cap\scr{C}$).

Then $\exists x \in \cap\scr{C}$ $: f(x) = y$.

Then $\forall C \in \scr{C}$ $: x \in C$.

Then $\forall C \in \scr{C}$, $y \in f(C)$.

Then $y \in \cap f(\scr{C})$.

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For the RHS $\Longrightarrow$ LHS, I proved it just going backwards. I haven't thought of a contradiction where this fails.

Let $y \in \cap f(\scr{C})$.

Then $\forall C \in \scr{C}$, $y \in f(C)$.

Then $ \exists x \in C \hspace{2 pt} [\forall C \in \scr{C}$ $: f(x) = y]$.

Then $x \in \cap \scr{C}$.

So $y \in f(\cap\scr{C})$.

Answer 1

I’ll take things in order. The second line of the first argument isn’t right: $f$ need not be injective, so $f^{-1}(y)$ need not even make sense. You could write $x\in f^{-1}[\{y\}]$, but it would be better and clearer to say simply that there is an $x\in\bigcap\mathscr{C}$ such that $f(x)=y$. The remainder of the first argument, however, is fine.

The second argument is simply wrong, and the conclusion of it is false. Specifically, the third line is false: for each $C\in\mathscr{C}$ there is an $x_C\in C$ such that $f(x_C)=y$, but it might be a different $x_C$ for each $C\in\mathscr{C}$. Here, as in the erroneous step in the first argument, you seem to be unconsciously assuming that $f$ is injective, so that a given $y\in Y$ can be the image of only one point of $X$.

To see specifically how this can happen, imagine that $X=\{0,1\}$, $Y=\{0\}$, $f(0)=f(1)=0$, and $\mathscr{C}=\{\{0\},\{1\}\}$. $\bigcap\mathscr{C}=\varnothing$, so $f\left(\bigcap\mathscr{C}\right)=\varnothing$, but $0\in\bigcap f\left(\mathscr{C}\right)$.

Answer 2

For the simplest case take two subsets $A$ and $B$ of a set $E$ and $f:E\longrightarrow F$ a map and then it is already true that $f(A\cap B)\subset f(A)\cap f(B)$ but not the converse. Here is a counterexample. Let $E=\{x_1,x_2\}$ and $F=\{y\}$ and $$f:E\rightarrow F;\; x_1\mapsto y\; \text{and}\; x_2\mapsto y$$

Let $A=\{x_1\}$ and $B=\{x_2\}$. We have that $f(A\cap B)=f(\emptyset)=\emptyset$ while $f(A)\cap f(B)=\{y\}\cap \{y\}=\{y\}$. Hence the result.