Example Where \(g \circ f\) is Injective but \(g\) is Not Injective
📺 Video Explanation
📝 Question
Give examples of two functions:
\[ f:\mathbb{N}\to\mathbb{Z},\qquad g:\mathbb{Z}\to\mathbb{Z} \]
such that:
\[ g\circ f \text{ is injective, but } g \text{ is not injective.} \]
✅ Solution
🔹 Choose Functions
Take:
\[ f(x)=x \]
and
\[ g(x)=|x| \]
Then:
\[ f:\mathbb{N}\to\mathbb{Z},\qquad g:\mathbb{Z}\to\mathbb{Z} \]
These satisfy the required condition. :contentReference[oaicite:1]{index=1}
🔹 Step 1: Show that \(g\) is not injective
For injective function, different inputs must give different outputs.
But:
\[ g(1)=|1|=1 \]
and
\[ g(-1)=|-1|=1 \]
So:
\[ g(1)=g(-1) \quad\text{but}\quad 1\ne -1 \]
Therefore:
\[ g \text{ is not injective} \]
🔹 Step 2: Find \(g\circ f\)
By definition:
\[ (g\circ f)(x)=g(f(x)) \]
Since:
\[ f(x)=x \]
So:
\[ (g\circ f)(x)=g(x)=|x| \]
🔹 Step 3: Show that \(g\circ f\) is injective
Let:
\[ (g\circ f)(x_1)=(g\circ f)(x_2) \]
Then:
\[ |x_1|=|x_2| \]
Since:
\[ x_1,x_2\in\mathbb{N} \]
Natural numbers are positive, so:
\[ |x_1|=x_1,\qquad |x_2|=x_2 \]
Thus:
\[ x_1=x_2 \]
Therefore:
\[ g\circ f \text{ is injective} \]
🎯 Final Answer
One suitable example is:
\[ \boxed{f(x)=x,\qquad g(x)=|x|} \]
Then:
\[ \boxed{(g\circ f)(x)=|x|} \]
So, \(g\circ f\) is injective but \(g\) is not injective. :contentReference[oaicite:2]{index=2}
🚀 Exam Shortcut
- Take \(f\) as identity on \(\mathbb{N}\)
- Take \(g\) as modulus on \(\mathbb{Z}\)
- \(|x|\) fails injective on integers but works on naturals