If functions f - A - B and g - B - A satisfy gof - I A- then show that f is one-one and g is onto-

Given that, f: A → B and g: B → A satisfy gof =I_A ∵ gof =I_A ⇒ gof{f(x_1)}=gof{f(x_2)} ⇒ g(x_1)=g(x_2) [∵ gof =I_A] ∴ x_1 =x_2 Hence, f is one one and g is ont ...

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Standard XII

Mathematics

Strictly Increasing Functions

If functions ...

Question

If functions $f : A \to B$ and $g : B \to A$ satisfy $g o f = I_{A}$ , then show that f is one-one and g is onto.

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Solution

Given that,
$f : A \to B$ and $g : B \to A$ satisfy $g o f = I_{A}$
$∵ g o f = I_{A} \Rightarrow g o f {f (x_{1})} = g o f {f (x_{2})} \Rightarrow g (x_{1}) = g (x_{2}) [∵ g o f = I_{A}] ∴ x_{1} = x_{2}$
Hence, f is one-one and g is onto.

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Similar questions

Q. If

f : A \to B

and

g : B \to C

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f : A \to B

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Q. Let

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If functions f:A→B and g:B→A satisfy gof=IA, then show that f is one-one and g is onto.

Given that, f:A→B and g:B→A satisfy gof=IA ∵gof=IA⇒gof{f(x1)}=gof{f(x2)}⇒g(x1)=g(x2)[∵gof=IA]∴x1=x2 Hence, f is one-one and g is onto.

If functions $f : A \to B$ and $g : B \to A$ satisfy $g o f = I_{A}$ , then show that f is one-one and g is onto.

Given that,
$f : A \to B$ and $g : B \to A$ satisfy $g o f = I_{A}$
$∵ g o f = I_{A} \Rightarrow g o f {f (x_{1})} = g o f {f (x_{2})} \Rightarrow g (x_{1}) = g (x_{2}) [∵ g o f = I_{A}] ∴ x_{1} = x_{2}$
Hence, f is one-one and g is onto.