prove $f\left(A∩\ f^{-1}\left(B\right)\right)=f\left(A\right)\ ∩\ B$ and $f\left(\ f^{-1}\left(B\right)\right)=f\left(X\right)\ ∩\ B$

Question

Assume $f:\ X→Y$ and $A⊆X\ ,\ B⊆Y$ then prove:

$$f\left(A∩\ f^{-1}\left(B\right)\right)=f\left(A\right)\ ∩\ B$$

$$f\left(\ f^{-1}\left(B\right)\right)=f\left(X\right)\ ∩\ B$$

For the first one assume $y\ ∈\ f\left(A∩\ f^{-1}\left(B\right)\right)$, iff there exist $x\ ∈\left(A∩\ f^{-1}\left(B\right)\right)$ such that $f(x)=y$ , in other words there exist $x$$\ ∈\ A$ such that $f(x)=y$ and there exist $x$$\ ∈f^{-1}\left(B\right)$ such that $f(x)=y$, implies $y\ ∈\ f\left(A\right)\ \text{and}\ f\left(x\right)=y\ ∈\ B$

For the other one let $y\ ∈\ f\left(\ f^{-1}\left(B\right)\right)$ this holds iff $\ ∃\ x\ ∈\ f^{-1}\left(B\right)$ such that $f\left(x\right)=y$ or equivalently $\ ∃f\left(x\right)\ ∈B$ such that $\ f\left(x\right)=y$ hence $\ y∈\ \ B$

Answer 1

$y$ in $f(A \cap f^{-1}(B))$ iff
exists $x$ in $A \cap f^{-1}(B))$ with $y = f(x)$ iff
exists x in A with $(f(x)$ in $B$ and $y = f(x))$ iff
$y$ in $f(A)$ and $y$ in $B$.

The second equation is an immediate corollary
of the first equation by setting $A$ to $X$.

Answer 2

You've only done one inclusion: $$f[A \cap f^{-1}[B]] \subseteq f[A]\cap B$$

(I use brackets $[]$ for the set (inverse) images of functions, and parentheses $()$ for evaluations of elements of the domain).

You also need the reverse inclusion for equality of sets: if $y \in f[A] \cap B$, so we know $y \in B$ and there is some $x \in A$ with $f(x)=y$. Then $x \in f^{-1}[B]$ by definition (its image is in $B$) so $x \in A \cap f^{-1}[B]$ and as $y=f(x)$, so we know $y \in f[A \cap f^{-1}[B]]$, as required.

Likewise your second proof shows

$$f[f^{-1}[B]] \subseteq f[X] \cap B$$ and we again also need the reverse inclusion: if $y \in f[X] \cap B$, so we know that $y=f(x)$ for some $x \in X$ and that $y \in B$, so by definition $x \in f^{-1}[B]$ and $y=f(x)$ shows that $y \in f[f^{-1}[B]]$, as required.