Let A and B be non-empty sets in mathbb{R} an | vedclass

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(D) function $f : A 	o B$ is bijective if it is both one-to-one (injective) and onto (surjective). Statement $1$ states that $f$ is an onto function,

Vedclass · Accepted Answer

Statement $1$ is true,Statement $2$ is true; Statement $2$ is not the correct explanation for Statement $1$.

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(D) function $f : A 	o B$ is bijective if it is both one-to-one (injective) and onto (surjective). Statement $1$ states that $f$ is an onto function,which is true by the definition of a bijective function. Statement $2$ states that there exists a function $g : B 	o A$ such that $f \circ g = I_B$. Since $f$ is bijective,it is invertible,meaning there exists an inverse function $f^{-1} : B 	o A$ such that $f \circ f^{-1} = I_B$. Thus,$g = f^{-1}$ satisfies the condition. Therefore,Statement $2$ is also true. However,Statement $2$ describes the property of invertibility resulting from bijectivity,while Statement $1$ is simply the definition of one part of bijectivity. Statement $2$ is not the reason why $f$ is onto; rather,both are consequences of $f$ being bijective. Thus,Statement $2$ is not the correct explanation for Statement $1$.

List-$I$	List-$II$
$(A) f(-5)+f(0)+f(-1)$	$(I) 16$
$(B) f(f(5)+10f(-3))$	$(II) 40$
$(C) f(f(-4))$	$(III) -31$
$(D) f(f(f(1)))$	$(IV) -12$
	$(V) 19$

Let $A$ and $B$ be non-empty sets in $\mathbb{R}$ and $f : A \to B$ be a bijective function.
Statement $1$ : $f$ is an onto function.
Statement $2$ : There exists a function $g : B \to A$ such that $f \circ g = I_B$.

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Let $A$ and $B$ be non-empty sets in $\mathbb{R}$ and $f : A \to B$ be a bijective function. Statement $1$ : $f$ is an onto function. Statement $2$ : There exists a function $g : B \to A$ such that $f \circ g = I_B$.