Question:
Consider function \(f:A\to B\) and \(g:B\to C\) (\(A,B,C\subseteq\mathbb{R}\)) such that \((gof)^{-1}\) exists, then :
Consider function \(f:A\to B\) and \(g:B\to C\) (\(A,B,C\subseteq\mathbb{R}\)) such that \((gof)^{-1}\) exists, then :
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\((gof)^{-1}\) exists iff \(gof\) is bijective; deduce what this forces on \(f\) and \(g\) separately.
Updated On: Nov 4, 2026
- \(f\) and \(g\) both are one-one
- \(f\) and \(g\) both are onto
- \(f\) is one-one and \(g\) is onto
- \(f\) is onto and \(g\) is one-one
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The Correct Option is C
Solution and Explanation
\((gof)^{-1}\) exists \(\Rightarrow gof:A\to C\) is one-one and onto.
If \(f\) were not one-one, \(f(x_1)=f(x_2)\) for some \(x_1\ne x_2\) would give \(gof(x_1)=gof(x_2)\), contradicting injectivity. So \(f\) is one-one.
Since \(gof\) is onto \(C\), every element of \(C\) is of the form \(g(f(x))\), so \(g\) is onto \(C\).
(Nothing forces \(f\) to be onto or \(g\) to be one-one.)
Answer: (3) \(f\) is one-one and \(g\) is onto
If \(f\) were not one-one, \(f(x_1)=f(x_2)\) for some \(x_1\ne x_2\) would give \(gof(x_1)=gof(x_2)\), contradicting injectivity. So \(f\) is one-one.
Since \(gof\) is onto \(C\), every element of \(C\) is of the form \(g(f(x))\), so \(g\) is onto \(C\).
(Nothing forces \(f\) to be onto or \(g\) to be one-one.)
Answer: (3) \(f\) is one-one and \(g\) is onto
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