Composition Proof I am trying to study functions in math and learning some basic proofs. In numerous places I have seen this: $$(f \circ\ g) ^{-1}(u) = g^{-1}(f^{-1}(u))$$
I know this is true as well, having used it in numerous places in middle and high school. Is there any way of proving this definition though using logical steps? Thank you!

Assuming that $g : A \to B$ and $f : B \to C$, then, by definition, $(f\circ g)^{-1} : C\to A$ is the unique function such that

*

*$(f\circ g)^{-1}\circ(f\circ g) = \textrm{id}_A$, and

*$(f\circ g)\circ(f\circ g)^{-1} = \textrm{id}_C$.

But observe that the function $g^{-1}\circ f^{-1} : C\to A$ has the same properties:
$$\begin{align}
(g^{-1}\circ f^{-1}) \circ (f\circ g) 
&= \big( g^{-1} \circ (f^{-1} \circ f) \big) \circ g \\
&= ( g^{-1} \circ \textrm{id}_B) \circ g \\
&= g^{-1} \circ g = \textrm{id}_A
\end{align}$$
and
$$\begin{align}
(f\circ g) \circ (g^{-1}\circ f^{-1})
&= \big( f \circ (g \circ g^{-1}) \big) \circ f^{-1} \\
&= ( f \circ \textrm{id}_B) \circ f^{-1} \\
&= f \circ f^{-1} = \textrm{id}_C.
\end{align}$$
Hence, $g^{-1}\circ f^{-1}$ and $(f\circ g)^{-1}$ are the same!
