Prove that \(f\circ g\) is Injection and Surjection

📝 Question

Let \(f:A\to A\) and \(g:A\to A\) be two bijections. Prove that:

(i) \(f\circ g\) is injective

(ii) \(f\circ g\) is surjective

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✅ Solution

🔹 Step 1: Given

Since \(f\) and \(g\) are bijections, they are both:

• Injective (one-one)
• Surjective (onto)

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🔹 Step 2: Prove \(f\circ g\) is injective

Assume:

\[ (f\circ g)(x_1)=(f\circ g)(x_2) \]

\[ f(g(x_1))=f(g(x_2)) \]

Since \(f\) is injective:

\[ g(x_1)=g(x_2) \]

Since \(g\) is injective:

\[ x_1=x_2 \]

Hence, \(f\circ g\) is injective.

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🔹 Step 3: Prove \(f\circ g\) is surjective

Let \(y \in A\).

Since \(f\) is surjective, there exists \(z \in A\) such that:

\[ f(z)=y \]

Since \(g\) is surjective, there exists \(x \in A\) such that:

\[ g(x)=z \]

Therefore:

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

Hence, \(f\circ g\) is surjective.

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🔹 Step 4: Conclusion

Since \(f\circ g\) is both injective and surjective, it is bijective.

This is a standard result: the composition of bijections is again a bijection. :contentReference[oaicite:0]{index=0}

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🎯 Final Answer

\[ \boxed{f\circ g \text{ is injective and surjective}} \] —

🚀 Exam Shortcut

• Injection: use injectivity of both \(f\) and \(g\)
• Surjection: use surjectivity step-by-step
• Composition of bijections ⇒ always bijection