Q.Let $A = \{0, 1\}$ and $\mathbb{N}$ be the set of natural numbers. Then the mapping $f : \mathbb{N} \to A$ defined by $f(2n - 1) = 0$, $f(2n) = 1$, $\forall\, n \in \mathbb{N}$, is onto.

Concept: Onto (surjective) function — every element of the codomain $A$ must have at least one preimage in $\mathbb{N}$.

Reasoning:

1. The codomain is $A = \{0, 1\}$.
2. For $0$: $f(2n - 1) = 0$ for all $n \in \mathbb{N}$. Taking $n = 1$ gives $f(1) = 0$, so $0$ has a preimage. …

The question asks whether the given mapping is onto (also called surjective). A function $f: X \to Y$ is onto if every element of the codomain $Y$ is the image of at least one element from the domain $X$. In simpler terms: nothing in $Y$ is left out — every possible output actually occurs.

Here, the codomain is $A = \{0, 1\}$, a tiny set with just two elements. The domain is $\mathbb{N}$, the set of natural numbers (usually $\{1,2,3,\dots\}$). The function is defined piecewise:

• For odd numbers: $f(2n-1) = 0$
• For even numbers: $f(2n) = 1$

Let’s check whether both $0$ and $1$ are actually hit.

1. Check for $0$:

   Take $n = 1$. Then $2n-1 = 1$, and $f(1) = 0$. So $0$ is achieved. (In fact, every odd natural number maps to $0$, so there are infinitely many preimages.)

2. Check for $1$:

   Take $n = 1$ again for the even case: $2n = 2$, and $f(2) = 1$. So $1$ is also achieved. (Every even natural number maps to $1$.)

3. Conclusion:

   Both elements of $A$ have at least one preimage in $\mathbb{N}$. Therefore, the function is onto. …

Method: Checking Whether a Function is Onto (Surjective)

Use this when you must decide if every element of the codomain is actually produced as an output.

Steps

Step 1: Identify the codomain clearly

Onto is a statement about the codomain, so list exactly which values must be "hit". Here the codomain is the finite set $A=\{0,1\}$.

Step 2: For each codomain element, find at least one preimage

Pick a target $y$ and solve $f(x)=y$ for some $x$ in the domain. Producing even one such $x$ means that value is covered.

Step 3: Handle a piecewise rule piece by piece …

Common Mistakes

Mistake 1: Confusing onto with one-one

Why it's wrong: onto only asks whether the codomain is fully covered; it does not care that many inputs (all odd numbers) map to the same output. Correct approach: for onto, check coverage of outputs, not uniqueness of inputs.

Mistake 2: Declaring it "not onto" because the function is many-to-one

Why it's wrong: repeated outputs are irrelevant to surjectivity. Correct approach: as long as both $0$ and $1$ are attained, the map is onto regardless of how many inputs share them. …