Answer
1. Since \(f\) and \(g\) are bijections, their composition \(g \circ f\) is also a bijection.
2. The inverse of \(g \circ f\) exists and is given by \(\,(g \circ f)^{-1} = f^{-1} \circ g^{-1}\).
Work Step by Step
We want to show two things:
1. **\((g \circ f)\) is a bijection** \(\bigl(X \to Z\bigr)\).
2. **Its inverse** is \(\displaystyle (g \circ f)^{-1} = f^{-1} \circ g^{-1}\).
Here, \(f : X \to Y\) and \(g : Y \to Z\) are each one-to‐one (injective) and onto (surjective), hence **bijections** with well‐defined inverses \(f^{-1}\) and \(g^{-1}\).
---
## 1. Show \(g \circ f\) is a Bijection
### (a) Injectivity
Suppose \((g \circ f)(x_1) = (g \circ f)(x_2)\). That is,
\[
g\bigl(f(x_1)\bigr) \;=\; g\bigl(f(x_2)\bigr).
\]
Because \(g\) is injective, we must have
\[
f(x_1) \;=\; f(x_2).
\]
Since \(f\) is also injective, it follows \(x_1 = x_2\). Thus \(g \circ f\) is injective.
### (b) Surjectivity
Take any \(z \in Z\). Because \(g\) is onto, there exists \(y \in Y\) such that \(g(y) = z\). Moreover, since \(f\) is onto, there exists \(x \in X\) with \(f(x) = y\). Then
\[
(g \circ f)(x)
\;=\;
g\bigl(f(x)\bigr)
\;=\;
g(y)
\;=\;
z.
\]
Hence every \(z \in Z\) is in the image of \((g \circ f)\). Thus \(g \circ f\) is surjective.
Combining injectivity and surjectivity shows \(g \circ f\) is a **bijection** \(X \to Z\).
---
## 2. Show \(\,(g \circ f)^{-1} = f^{-1} \circ g^{-1}\)
To prove two functions \((g \circ f)^{-1}\) and \(\bigl(f^{-1} \circ g^{-1}\bigr)\) are equal, we can check that each “undoes” the action of \(g \circ f\) from both sides.
### (a) \((g \circ f)\bigl(f^{-1} \circ g^{-1}(z)\bigr) = z\)
Let \(z \in Z\). Define
\[
x \;=\; \bigl(f^{-1} \circ g^{-1}\bigr)(z).
\]
Then \(g^{-1}(z) \in Y\) and \(f^{-1}\bigl(g^{-1}(z)\bigr) \in X\). We compute:
\[
(g \circ f)(x)
\;=\;
g\bigl(f(x)\bigr)
\;=\;
g\Bigl(f\bigl(f^{-1}\bigl(g^{-1}(z)\bigr)\bigr)\Bigr)
\;=\;
g\bigl(g^{-1}(z)\bigr)
\;=\;
z.
\]
Hence for every \(z\), \((g \circ f)\bigl(\dots\bigl)=z\). This shows that \(f^{-1}\circ g^{-1}\) is a right‐inverse of \(g \circ f\).
### (b) \(\bigl(f^{-1} \circ g^{-1}\bigr)\bigl((g \circ f)(x)\bigr) = x\)
Let \(x \in X\). Then
\[
\bigl(f^{-1} \circ g^{-1}\bigr)\bigl((g \circ f)(x)\bigr)
\;=\;
f^{-1}\Bigl(g^{-1}\bigl(g\bigl(f(x)\bigr)\bigr)\Bigr)
\;=\;
f^{-1}\bigl(f(x)\bigr)
\;=\;
x.
\]
So \(f^{-1}\circ g^{-1}\) is also a left‐inverse of \(g \circ f\).
Together, (a) and (b) show that \(\bigl(f^{-1} \circ g^{-1}\bigr)\) is indeed the two‐sided inverse of \(g \circ f\). Thus:
\[
\boxed{
(g \circ f)^{-1}
\;=\;
f^{-1} \;\circ\; g^{-1}.
}
\]
