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

You can also prove it by contradiction.
Suppose otherwise that $y\in f(f^{-1}(D))$ and $y\not\in D$. According to the definitions involved, we get
\begin{align*}
y\in f(f^{-1}(D)) & \Rightarrow (\exists x\in f^{-1}(D))(y = f(x))
\end{align*}
But we also know that $x\in f^{-1}(D)$ iff $f(x)\in D$.
Hence we obtain a contradiction and the desired result holds.
Hopefully this helps!
