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!
 A: 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!
A: Assume functions $g:A\mapsto B, f:B\mapsto C$ are bijective (therefore invertible).
By definition : $(f\circ g)(x)=f(g(x))$ 
Applying inversion:
$$g^{-1}(f^{-1}((f\circ g)(x)))=x$$
Let : $x=(f\circ g)^{-1}(u)$ which means also $u=(f\circ g)(x)$
Therefore …$$g^{-1}(f^{-1}(u))=(f\circ g)^{-1}(u)$$
A: To prove something is the inverse of a function, you need only check the compositions work as expected, i.e. if you believe $g(x)$ is the inverse of $f(x)$, then show $f(g(x))=x$ and $g(f(x))=x$. This will prove that $g(x)= f^{-1}(x)$.
Define $G(u)= (f \circ g)(u)$ and $H(u)= g^{-1}(f^{-1}(u))$. Now just check that $H(G(u))=u$ and $G(H(u))=u$. You will want to use the fact that $f,g$ have inverses, i.e. $f^{-1}, g^{-1}$ exist so that $f(f^{-1}(x))=x$, $f^{-1}(f(x))=x$ and $g(g^{-1}(x))=x$, $g^{-1}(g(x))=x$.
A: For any $x$ in the domain, let $y=(f\circ g)^{-1}(x)$. Note that the following are equivalent:
$$y=(f\circ g)^{-1}(x)$$
$$(f\circ g)(y)=x$$
$$f(g(y))=x$$
$$g(y)=f^{-1}(x)$$
$$y=g^{-1}(f^{-1}(x))$$
$$y=(g^{-1}\circ f^{-1})(x)$$
It follows that $(f\circ g)^{-1}(x)=y=(g^{-1}\circ f^{-1})(x)$
