direct image proof $f(f^{-1}(D)) \subseteq D$ Suppose that $f\colon A \rightarrow B$ is a function, and $C \subseteq A$ and $D \subseteq B$ prove that $f(f^{-1}(D)) \subseteq D$
I got this
let $z\in  f(f^{-1}(D)) \Rightarrow (\exists x)(x \in f^{-1}(D)$  and  $z=f(x))$
$ \Rightarrow (\exists x)(f(x)\in D$  and $z=f(x))$
$ \Rightarrow (\exists x)(z\in D$  and $z=f(x))$
I do not know if I am okay or not, and as well I do not know how to conclude

It might help for you to define the sets $f(f^{-1}(D))$ and $f^{-1}(D)$ before you start; for me, at least, this makes the proof easier. It also saves the writing you need to do:
$$f(f^{-1}(D))=\{f(x):x \in f^{-1}(D)\}$$
$$f^{-1}(D)=\{a \in A:f(a) \in D \}$$
Now you start the proof. Suppose $z \in f(f^{-1}(D))$. Then there exists $x \in f^{-1}(D)$ such that $f(x)=z$. But if $x \in f^{-1}(D)$, then $z=f(x)\in D$. So $f(f^{-1}(D)) \subseteq D$, and this is what we had to prove.
