Verify \(f\) and \(g\) are Inverse Functions Using \(f \circ g\) and \(g \circ f\)

📺 Video Explanation

📝 Question

Let

\[ A=\{a,b,c\},\qquad B=\{u,v,w\} \]

Let:

\[ f=\{(a,v),(b,u),(c,w)\} \]

and

\[ g=\{(u,b),(v,a),(w,c)\} \]

where \(f:A\to B\) and \(g:B\to A\). Verify whether \(f\) and \(g\) are inverse functions.

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✅ Solution

🔹 Step 1: Find \(g\circ f\)

By definition:

\[ (g\circ f)(x)=g(f(x)) \]

• \(f(a)=v \Rightarrow g(v)=a\)
• \(f(b)=u \Rightarrow g(u)=b\)
• \(f(c)=w \Rightarrow g(w)=c\)

So:

\[ g\circ f=\{(a,a),(b,b),(c,c)\} \]

Thus, \(g\circ f=I_A\), the identity function on set \(A\). :contentReference[oaicite:1]{index=1}

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🔹 Step 2: Find \(f\circ g\)

By definition:

\[ (f\circ g)(x)=f(g(x)) \]

• \(g(u)=b \Rightarrow f(b)=u\)
• \(g(v)=a \Rightarrow f(a)=v\)
• \(g(w)=c \Rightarrow f(c)=w\)

So:

\[ f\circ g=\{(u,u),(v,v),(w,w)\} \]

Thus, \(f\circ g=I_B\), the identity function on set \(B\). :contentReference[oaicite:2]{index=2}

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🎯 Final Answer

\[ \boxed{g\circ f=I_A=\{(a,a),(b,b),(c,c)\}} \]

\[ \boxed{f\circ g=I_B=\{(u,u),(v,v),(w,w)\}} \]

Hence, \(f\) and \(g\) are inverse functions of each other.

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🚀 Exam Shortcut

• If both compositions give identity, functions are inverses
• \(g\circ f\) should return original elements of set \(A\)
• \(f\circ g\) should return original elements of set \(B\)