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?

  • $\begingroup$ How do you know $y\in A$ in the last line? You only have, from your definitions that $y\in X$. $\endgroup$ – Michael Burr Jan 29 '17 at 20:03
  • $\begingroup$ Alright i see. I think that is my problem. How do i correct it? $\endgroup$ – Mathe Jan 29 '17 at 20:08

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 $A$.

  • $\begingroup$ Proof that: $f(f^{-1}(a)) \subseteq A:$ Let $y \in f(f^{-1}(A)$ If we can prove that $y \in A$ we would know that $f(f^{-1}(A) \subseteq A$ By definition: $f^{-1}(A)=\{x \in X\mid f(x) \in A\}$ Therefore: $x \in f^{-1}(A): f(x)=y \ \wedge \ f(x) \in A $\\ $\Rightarrow y \in A$ $\endgroup$ – Mathe Jan 29 '17 at 21:05
  • $\begingroup$ Thanks for great response. I have tried to build on, what you showed me. Is this the correct way to formulate it? or did i miss something? $\endgroup$ – Mathe Jan 29 '17 at 21:06
  • $\begingroup$ @Mathe It comes down to the same thing. You do need a quantifier $\exists x$ if you're using formulae. I prefer words above formulae, though. They express the ideas just as well. TO show the inclusion, you have to start with an arbitary $y$ in the left hand side, which you then introduce. Don't let variables drop out of thin air! $\endgroup$ – Henno Brandsma Jan 29 '17 at 21:19

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 :

  • $A$ lies in 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 $f : X\to Y$
  • $B$ lies in 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

$$f(B)=\{f(x) \mid x \in B\} = \{y \in X \mid \exists x \in B: y=f(x)\}$$

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

  • $\begingroup$ It is interesting to note that, equality holds when f is surjective. $\endgroup$ – math is love Jan 29 '17 at 20:30

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 $\{f(x)\mid x\in f^{-1}(A) \}\subset A$.


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 $f(f^{-1}(A))\subset A$


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