Show \(g \circ f\) is One-One if \(f\) and \(g\) are One-One Functions
📺 Video Explanation
📝 Question
If:
\[ f:A\to B \quad \text{and} \quad g:B\to C \]
are one-one (injective) functions, show that:
\[ g\circ f:A\to C \]
is also a one-one function.
✅ Solution
🔹 Step 1: Use definition of one-one function
A function is one-one if:
\[ F(x_1)=F(x_2)\Rightarrow x_1=x_2 \]
We must prove this for:
\[ g\circ f \]
🔹 Step 2: Assume equal outputs
Let:
\[ (g\circ f)(x_1)=(g\circ f)(x_2) \]
By composition:
\[ g(f(x_1))=g(f(x_2)) \]
🔹 Step 3: Use injectivity of \(g\)
Since \(g\) is one-one:
\[ g(f(x_1))=g(f(x_2)) \Rightarrow f(x_1)=f(x_2) \]
🔹 Step 4: Use injectivity of \(f\)
Since \(f\) is one-one:
\[ f(x_1)=f(x_2)\Rightarrow x_1=x_2 \]
🎯 Final Answer
Thus:
\[ (g\circ f)(x_1)=(g\circ f)(x_2)\Rightarrow x_1=x_2 \]
Therefore:
\[ \boxed{g\circ f \text{ is a one-one function}} \]
Hence, the composition of two injective functions is also injective. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- Equal outputs of composition ⇒ equal outputs under outer function
- Use one-one property of outer function first
- Then use one-one property of inner function