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

The problem gives a bijective map $$f : A \rightarrow B$$ between two non-empty subsets of $$\mathbb{R}$$ and asks us to compare two statements.

Statement 1: “$$f$$ is an onto (surjective) function.”
Statement 2: “There exists a function $$g : B \rightarrow A$$ such that $$f \circ g = I_B$$ (the identity map on $$B$$).”

Recall the definitions first.

• A function is called bijective when it is both one-one (injective) and onto (surjective).
• A function $$g : B \rightarrow A$$ satisfying $$f\circ g = I_B$$ is called a right inverse of $$f$$. A right inverse exists precisely when $$f$$ is surjective.

Now check the two statements one by one.

Step 1: Truth of Statement 1
Since $$f$$ is given to be bijective, surjectivity is already included in the definition. Therefore $$f$$ is indeed onto. Statement 1 is true.

Step 2: Truth of Statement 2
For each $$y\in B$$, surjectivity ensures the existence of at least one $$x\in A$$ with $$f(x)=y$$. Choose one such $$x$$ (using the axiom of choice if necessary) and define $$g(y)=x$$. Then $$f(g(y)) = y$$ for every $$y\in B$$, i.e. $$f\circ g = I_B$$. Hence a right inverse $$g$$ exists and Statement 2 is also true.

Step 3: Does Statement 2 explain Statement 1?
Statement 1 is true solely because “bijective” already includes “onto.” While Statement 2 is a separate characterisation of surjectivity, it is not used as the reason in Statement 1; the onto property follows directly from the definition of bijection, not from the existence of a right inverse highlighted in Statement 2. Therefore Statement 2 is not the correct explanation for Statement 1.

Hence the correct option is:

Option D which is: Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation for Statement 1.