Show \(g \circ f\) is Onto if \(f\) and \(g\) are Onto Functions
📺 Video Explanation
📝 Question
If:
\[ f:A\to B \quad \text{and} \quad g:B\to C \]
are onto (surjective) functions, show that:
\[ g\circ f:A\to C \]
is also an onto function.
✅ Solution
🔹 Step 1: Use definition of onto function
A function is onto if every element of codomain has at least one pre-image.
We must prove that:
\[ \forall c\in C,\ \exists a\in A \text{ such that } (g\circ f)(a)=c \]
🔹 Step 2: Take any element in codomain \(C\)
Let:
\[ c\in C \]
Since \(g:B\to C\) is onto, there exists:
\[ b\in B \text{ such that } g(b)=c \]
🔹 Step 3: Use onto property of \(f\)
Since \(f:A\to B\) is onto, for this element \(b\in B\), there exists:
\[ a\in A \text{ such that } f(a)=b \]
🔹 Step 4: Find image under composition
Now:
\[ (g\circ f)(a)=g(f(a)) \]
Substitute \(f(a)=b\):
\[ (g\circ f)(a)=g(b) \]
And since:
\[ g(b)=c \]
Therefore:
\[ (g\circ f)(a)=c \]
🎯 Final Answer
For every:
\[ c\in C \]
there exists:
\[ a\in A \]
such that:
\[ (g\circ f)(a)=c \]
Therefore:
\[ \boxed{g\circ f \text{ is an onto function}} \]
Hence, the composition of two onto functions is also onto. :contentReference[oaicite:1]{index=1}
🚀 Exam Shortcut
- Pick any output in final codomain
- Use onto property of outer function first
- Then use onto property of inner function