Need help for proving that: $f(f^{-1}(A)) ⊆ A$ I realised I needed to show more information, which I now did:

$$f: X \rightarrow X \,\,\,\rm{and}\,\,\,\, A \subseteq X$$
Proof that: $$f(f^{-1}(A)) \subseteq A$$

This is my proof:
By defintion: 
$$f^{-1}(A)=\{x \in X\mid f(x) \in A\}$$ 
and
$$f(A)=\{f(x) \mid x \in A\} = \{y \in X \mid \exists x \in A: y=f(x)\} \subseteq X$$
Therefore we can end the proof by a final definition:\
$$f(f^{-1}(A))=\{y \in A: \exists x \in f^{-1}(A):y=f(x)\} \subseteq A$$
Is this a legit "proof"? And is it even a proof, when i only use definitions?
 A: If $y \in f[f^{-1}[A]]$, this means that there is an $x \in f^{-1}[A]$ such that $f(x) = y$. But $x \in f^{-1}[A]$ means that $f(x) \in A$, so $y \in A$, which shows the inclusion. We just apply the two definitions you have given.
Also in words: $f^{-1}[A]$ are all points that are mapped by $f$ into $A$. So its image under $f$ is a subset of $A$.
A: In fact, one can prove that $f(f^{-1}(A))=f(X)\cap A$, then $f(f^{-1}(A))\subseteq A$.
Proof: Let $y\in f(X)\cap A$, then it exists $x\in X$ such that $f(x)=y$ and $x\in f^{-1}(A)$ hence $y\in f(f^{-1}(A))$. Conversely let $y\in f(f^{-1}(A))$ then $y\in A$ and $y\in f(f^{-1}(A))\subset f(X)$ thus $y\in f(X)\cap A$.
About your proof: In order not to be confused you should use a "policy of variables" like :

*

* $A$ is a subset of the codomain of $f$ (what is confusing is that, in this exercise $X=dom(f)=codom(f)$). In fact, equality $f(f^{-1}(A))=f(X)\cap A$ holds in general, for $f : X\to Y$

* $B$ is a subset of the domain of $f$

* $x$ is in the domain and $y$ in the codomain 

in this respect, I would rewrite the last part of your solution as

$$f(B)=\{f(x) \mid x \in B\} = \{y \in Y \mid \exists x \in B: y=f(x)\}$$
Therefore we can end the proof by a final definition:
$$f(f^{-1}(A))=\{y \in Y: \exists x \in f^{-1}(A):y=f(x)\} \subseteq A$$

A: Note that $f(f^{-1}(A)) = \{f(x)\mid x\in f^{-1}(A) \}$. Now for $x\in f^{-1}(A)$ we have $f(x)\in A$ by definition. So $\{f(x)\mid x\in f^{-1}(A) \}\subset A$.
A: We just have to follow the definitions of function, subset, range, etc. See:
If $y\in f(f^{-1}(A))$, then there exists $x\in f^{-1}(A)$ such that $f(x)=y$. Since $x\in f^{-1}(A)$, there exists $y'\in A$ such that $f(x)=y'$.
Note that $y=y'$ by definition. Since $y'$ is in $A$, we have $y\in A$
Therefore, we have just proved $y\in f(f^{-1}(A))\implies y\in A$, which is equivalent to prove $f(f^{-1}(A))\subset A$
