Show \(f\), \(g\), and \(g \circ f\) are Invertible and Verify Inverse Rule
📺 Video Explanation
📝 Question
Let:
\[ f:\{1,2,3\}\to\{a,b,c\} \]
defined by:
\[ f(1)=a,\quad f(2)=b,\quad f(3)=c \]
and:
\[ g:\{a,b,c\}\to\{\text{apple, ball, cat}\} \]
defined by:
\[ g(a)=\text{apple},\quad g(b)=\text{ball},\quad g(c)=\text{cat} \]
Show that \(f\), \(g\), and \(g\circ f\) are invertible. Find:
- \(f^{-1}\)
- \(g^{-1}\)
- \((g\circ f)^{-1}\)
Also verify:
\[ (g\circ f)^{-1}=f^{-1}\circ g^{-1} \]
✅ Solution
🔹 Step 1: Show \(f\) is invertible
Function \(f\) maps:
\[ 1\to a,\quad 2\to b,\quad 3\to c \]
All images are distinct and every element of codomain is used.
So \(f\) is bijective.
Hence inverse exists:
\[ \boxed{f^{-1}=\{(a,1),(b,2),(c,3)\}} \]
🔹 Step 2: Show \(g\) is invertible
Function \(g\) maps:
\[ a\to apple,\quad b\to ball,\quad c\to cat \]
All outputs are distinct and every codomain element is covered.
So \(g\) is bijective.
Hence:
\[ \boxed{ g^{-1}= \{ (\text{apple},a), (\text{ball},b), (\text{cat},c) \} } \]
🔹 Step 3: Find \(g\circ f\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
So:
\[ (g\circ f)(1)=g(a)=\text{apple} \]
\[ (g\circ f)(2)=g(b)=\text{ball} \]
\[ (g\circ f)(3)=g(c)=\text{cat} \]
Thus:
\[ g\circ f= \{ (1,\text{apple}), (2,\text{ball}), (3,\text{cat}) \} \]
This is also bijective.
Therefore:
\[ \boxed{ (g\circ f)^{-1}= \{ (\text{apple},1), (\text{ball},2), (\text{cat},3) \} } \]
🔹 Step 4: Verify \((g\circ f)^{-1}=f^{-1}\circ g^{-1}\)
Now:
\[ f^{-1}\circ g^{-1} \]
Apply \(g^{-1}\) first:
- \(\text{apple}\to a\)
- \(\text{ball}\to b\)
- \(\text{cat}\to c\)
Then apply \(f^{-1}\):
- \(a\to1\)
- \(b\to2\)
- \(c\to3\)
So:
\[ f^{-1}\circ g^{-1} = \{ (\text{apple},1), (\text{ball},2), (\text{cat},3) \} \]
This equals:
\[ (g\circ f)^{-1} \]
🎯 Final Answer
\[ \boxed{f^{-1}=\{(a,1),(b,2),(c,3)\}} \]
\[ \boxed{ g^{-1}= \{ (\text{apple},a), (\text{ball},b), (\text{cat},c) \} } \]
\[ \boxed{ (g\circ f)^{-1}= \{ (\text{apple},1), (\text{ball},2), (\text{cat},3) \} } \]
And:
\[ \boxed{(g\circ f)^{-1}=f^{-1}\circ g^{-1}} \]
🚀 Exam Shortcut
- Inverse exists only for bijection
- Reverse ordered pairs to get inverse
- For composition inverse: reverse order of inverses