$f: X \to Y$ prove that if $B \subseteq Y$, then the image of the pre-image of $B$ is a subset of B, shown by these symbols respectively: $f[f^{-1}[A]]$
However, I don't see how this is true, because it was previously proven that: $A \subseteq f^{-1}[f[A]]$ where $A \subseteq X$
If that's true, let $B = f[A]$ and let $A \subseteq f^{-1}[B] = C$, so $f[f^{-1}[B]] \nsubseteq B$ which is obvious because C is a superset of A.
What's the problem?
Best Answer
For simplicity, consider the zero-function $f:\mathbb R\to\{0\}$ that sends everything to zero. The we can construct an example following your suggested contradictory pattern as follows:
Now clearly $A=[0,1]\subseteq C=\mathbb R$, still $f(C)=\{0\}=B$. There is just no problem there.
In general, when you define $B=f(A)$ you can be certain that every element of $B$ is in the image of $f$ and therefore has a non-empty pre-image. Hence $f(f^{-1}(B))=B$ and not just $f(f^{-1}(B))\subseteq B$ for all such examples.
Regarding the actual proof of the proposition, the pre-image is defined as follows: $$ f^{-1}(B)=\{x\in X\ \mid\ f(x)\in B\} $$ so clearly for any $x\in f^{-1}(B)$ we must have $f(x)\in B$. Hence $$ f(f^{-1}(B))\subseteq B $$