Let's say that $$f(x): \mathbb {R} \to \mathbb {R}, \quad f(x)=-x^2-1$$ and $A$ and $B$ are subsets of $\mathbb {R}$. I want to prove that $f(A∩B)=f(A)∩f(B)$ is a false statement.
For this, I changed the formula to $f(A)=-A^2-1$ and $f(B)=-B^2-1$. If $A=1$ then $f(A)=-2$ and if $B=2$ then $f(B)=-5$, so $f(A) \cap f(B)= \emptyset$. But doesn't $f(A \cap B)=\emptyset$ as well? I'm kind of stuck on this proof.
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Take $f: \mathbb{R} \to \mathbb{R}: x \mapsto 0$.
Then $$f(\{0\}\cap \{1\}) = f(\emptyset) = \emptyset$$
and $$f(\{0\}) \cap f(\{1\}) = \{0\} \cap \{0\} = \{0\}$$
In fact, for any function $f: X \to Y$ it is true that for $A, B \subseteq X$, $f(A \cap B) \subseteq f(A ) \cap f(B)$, but the converse can fail as you can see.
Note that the converse is true if $f$ is injective, which I leave as an instructive exercise for you.
Pick $$ f:\{a,b\}\to \{c\}; a\mapsto c, b\mapsto c. $$ Then $f(\{a\})\cap f(\{b\}) = \{c\}\cap \{c\}=\{c\}$. However, $f(\{a\}\cap \{b\}) = f(\emptyset)=\emptyset.$
In your case, analogously, you can take and constant function $f:\mathbb{R}\to \mathbb{R};x\mapsto c$ and then test two disjoint sets.