$f(A\cap B)=f(A)\cap f(B)$. Where's the mistake?

Question

I'm trying to prove something that is false, to see where is the contradiction. I want to prove that if $f:X\longrightarrow Y$ and $A,B\subseteq X$ then $f(A\cap B)=f(A)\cap f(B)$.

So, let $y \in f(A\cap B)$ then: \begin{align} &\Longrightarrow y=f(x):x\in A\cap B\\ &\Longrightarrow y=f(x):x\in A \wedge y=f(x):x\in B\\ &\Longrightarrow y\in f(A) \wedge y\in f(B)\\ &\Longrightarrow y \in f(A)\cap f(B)\\ &\therefore f(A\cap B)\subseteq f(A)\cap f(B). \end{align}

Also, if $y\in f(A)\cap f(B)$ then: \begin{align} &\Longrightarrow y\in f(A) \wedge y\in f(B)\\ &\Longrightarrow y=f(x):x\in A \wedge y=f(x):x\in B\\ &\Longrightarrow y=f(x):x\in A \wedge x \in B\tag{*}\\ &\Longrightarrow y=f(x):x\in A\cap B\\ &\Longrightarrow y\in f(A\cap B)\\ &\therefore f(A)\cap f(B) \subseteq f(A\cap B). \end{align}

Therefore the equality should be true. I don't see where is my mistake in the first block, in the second one I think the mistake is in the equation I signaled with $(*)$ but don't know what I'm violating.

For instance if \begin{align} f:X=\mathbb R\longrightarrow& Y=\mathbb R\\ x\longmapsto&1 \end{align} and $A=[0,1], B=[2,3]$ then $A\cap B=\emptyset$ and $f(A \cap B)=f(\emptyset)=\emptyset\neq \{1\}\cap\{1\}=\{1\}=f(A)\cap f(B)$ is a counterexample.

Answer 1

If $y\in f(A)\cap f(B)$ therefore $y=f(x)$ for a certain $x\in A$ and $y=f(u)$ for a certain $u\in B$, but $x\neq u$ a priori. To have the equality, you need the injectivity. Indded, if $f$ is injective, $$y=f(x)=f(u)\implies x=u.$$ But if $f$ is not injective, take for example $f(x)=x^2$, $A=[-1,0]$ and $B=[0,1]$. You have that $f(A)\cap f(B)=[0,1]$ but $$f(A\cap B)=\{0\}\subset f(A)\cap f(B).$$