Proof verification: $f$ injective $\implies \forall A,B \subseteq M:f(A \cap B)=f(A) \cap f(B)$

Question

Show that for $f: M \to N$, it holds that $$f \text{ injective} \implies \forall A,B \subseteq M:f(A \cap B)=f(A) \cap f(B).$$

What I did:

Let $y \in f(A\cap B)$. Since $f$ is injective, there exists a left inverse, so one can say that $$\begin{align} y \in f(A\cap B) &\iff \\ f^{-1}(y) \in A\cap B &\iff \\ f^{-1}(x) \in A \land f^{-1}(y)\in B &\iff \\ y \in f(A) \land y \in f(B) &\iff \\ y \in f(A) \cap f(B). \end{align}$$

Is this proof right?

Answer 1

If $g\colon M\to N$ is a left inverse, i.e,, $g(f(x))=x$ for all $x\in M$, and $C\subseteq M$, then $$ y\in f(C)\iff g(y)\in C$$ (as used in your first step) need not be true. In fact you only have "$\implies$" in general

Answer 2

You can try to show that if you take $y\in f(A\cap B)\rightarrow y \in f(A) \cap f(B)$ so $f(A\cap B) \subseteq f(A) \cap f(B)$ . Then showing $ y\in f(A) \cap f(B)\rightarrow y\in f(A\cap B) $ you can state $f(A) \cap f(B)\subseteq f(A\cap B) $.