There's a nice proof of the theorem about composite functions (see theorem 5 here) that states

$$(g\circ f)^{-1}=f^{-1}( g^{-1})$$

Notice that $f^{-1}$ means the inverse of $f$. Could anyone help me with proving similar equality for preimages?

The preimage of a function is defined as $f^{-1}[Y]=\{x \in X | f(x) \in Y\}$.

Let $f: X \to Y$ and $g: Y \to Z$. The theorem states that for every subset $S \subseteq Z$:

$$(g\circ f)^{-1}[S]=f^{-1}(g^{-1}[S])$$

So the thing is to show that two sets above are equal.

My attempt would be:

Let $M=(g\circ f)^{-1}[S] = \{x\in X | g(f(x))\in S\}$.

then $g^{-1}[S]$ is a set $G=\{y\in Y | g(y))\in S\}$, and $f^{-1}(g^{-1}[S])=\{x\in X | f(x) \in G\}$.

I'm not sure how to proceed with it next.


1 Answer 1


In general: $$x\in h^{-1}(S)\iff h(x)\in S$$ So the following statements are equivalent:

  • $x\in(g\circ f)^{-1}(S)$
  • $(g\circ f)(x)\in S$
  • $g(f(x))\in S$
  • $f(x)\in g^{-1}(S)$
  • $x\in f^{-1}(g^{-1}(S))$

This shows that $$(g\circ f)^{-1}(S)=f^{-1}(g^{-1}(S))$$

  • $\begingroup$ Do you know where I can find a proof of the first statement ($x\in h^{-1}(S)\iff h(x)\in S$)? $\endgroup$
    – FD_bfa
    Commented Jan 5, 2023 at 9:12
  • 2
    $\begingroup$ @FD_bfa By definition we have $h^{-1}(S):=\{x\in X\mid h(x)\in S\}$. This gives immediately: $$x\in h^{-1}(S)\iff h(x)\in S$$For completeness let me state that $h^{-1}$ is in this context not the notation of an inverse of $h$ (often there is confusion about that). $\endgroup$
    – drhab
    Commented Jan 5, 2023 at 9:17

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