Answer
1. \(\displaystyle g\circ f : a\mapsto w,\; b\mapsto v,\; c\mapsto u.\)
2. \(\displaystyle (g\circ f)^{-1}: w\mapsto a,\; v\mapsto b,\; u\mapsto c.\)
3. \(\displaystyle g^{-1}: v\mapsto x,\; w\mapsto y,\; u\mapsto z.\)
4. \(\displaystyle f^{-1}: y\mapsto a,\; x\mapsto b,\; z\mapsto c.\)
5. \(\displaystyle f^{-1}\circ g^{-1}: u\mapsto c,\; v\mapsto b,\; w\mapsto a.\)
6. \(\displaystyle (g\circ f)^{-1} = f^{-1}\circ g^{-1}.\)
Work Step by Step
Below is a step‐by‐step solution following the given definitions:
\(X = \{\,a,b,c\}\)
\(Y = \{\,x,y,z\}\)
\(Z = \{\,u,v,w\}\)
Functions:
1. \(f : X \to Y\)
\[
f(a)=y,\quad f(b)=x,\quad f(c)=z.
\]
2. \(g : Y \to Z\)
\[
g(x)=v,\quad g(y)=w,\quad g(z)=u.
\]
We want to find:
1. \(g \circ f\)
2. \((g \circ f)^{-1}\)
3. \(g^{-1}\)
4. \(f^{-1}\)
5. \(f^{-1}\circ g^{-1}\)
6. The relationship between \((g \circ f)^{-1}\) and \(f^{-1}\circ g^{-1}\).
---
## 1) Compute \(g \circ f\)
\[
(g \circ f)(x) \;=\; g\bigl(f(x)\bigr).
\]
So for each element of \(X\):
\((g \circ f)(a) = g\bigl(f(a)\bigr) = g(y) = w.\)
\((g \circ f)(b) = g\bigl(f(b)\bigr) = g(x) = v.\)
\((g \circ f)(c) = g\bigl(f(c)\bigr) = g(z) = u.\)
Hence
\[
g \circ f : X \;\to\; Z
\quad\text{is given by}\quad
a \mapsto w,\; b \mapsto v,\; c \mapsto u.
\]
---
## 2) Find \((g \circ f)^{-1}\)
Since \(g \circ f\) is a bijection from \(X\) onto \(Z\), its inverse is a map \(Z\to X\). From
\[
(g \circ f)(a) = w,\quad
(g \circ f)(b) = v,\quad
(g \circ f)(c) = u,
\]
we invert these pairs:
\(w \mapsto a\)
\(v \mapsto b\)
\(u \mapsto c\)
Therefore,
\[
(g \circ f)^{-1}(w) = a,\quad
(g \circ f)^{-1}(v) = b,\quad
(g \circ f)^{-1}(u) = c.
\]
---
## 3) Find \(g^{-1}\)
We know \(g\colon x\mapsto v,\;y\mapsto w,\;z\mapsto u\). Invert these:
\(v \mapsto x\)
\(w \mapsto y\)
\(u \mapsto z\)
Hence
\[
g^{-1} : Z \;\to\; Y,\quad
g^{-1}(v) = x,\quad g^{-1}(w)=y,\quad g^{-1}(u)=z.
\]
---
## 4) Find \(f^{-1}\)
Similarly, \(f\colon a\mapsto y,\; b\mapsto x,\; c\mapsto z\). Invert these:
\(y \mapsto a\)
\(x \mapsto b\)
\(z \mapsto c\)
Hence
\[
f^{-1} : Y \;\to\; X,\quad
f^{-1}(y)=a,\quad f^{-1}(x)=b,\quad f^{-1}(z)=c.
\]
---
## 5) Compute \(f^{-1}\circ g^{-1}\)
This is a map \(Z \to X\). For each element of \(Z\), apply \(g^{-1}\) first, then \(f^{-1}\):
1. \(u \overset{g^{-1}}{\longmapsto} z \overset{f^{-1}}{\longmapsto} c.\)
2. \(v \overset{g^{-1}}{\longmapsto} x \overset{f^{-1}}{\longmapsto} b.\)
3. \(w \overset{g^{-1}}{\longmapsto} y \overset{f^{-1}}{\longmapsto} a.\)
So
\[
(f^{-1}\circ g^{-1})(u)=c,\quad
(f^{-1}\circ g^{-1})(v)=b,\quad
(f^{-1}\circ g^{-1})(w)=a.
\]
---
## 6) Relationship of \((g \circ f)^{-1}\) and \(f^{-1}\circ g^{-1}\)
Comparing the results of Step 2 and Step 5:
\((g \circ f)^{-1}(u)=c\) matches \((f^{-1}\circ g^{-1})(u)=c\).
\((g \circ f)^{-1}(v)=b\) matches \((f^{-1}\circ g^{-1})(v)=b\).
\((g \circ f)^{-1}(w)=a\) matches \((f^{-1}\circ g^{-1})(w)=a\).
Thus **they are the same function**. Indeed, a standard theorem says:
\[
\boxed{(g \circ f)^{-1} \;=\; f^{-1} \;\circ\; g^{-1}.}
\]
