Q.Consider the identity function $I_{\mathbb{N}}: \mathbb{N} \to \mathbb{N}$ defined as $I_{\mathbb{N}}(x) = x \; \forall x \in \mathbb{N}$. Show that although $I_{\mathbb{N}}$ is onto but $I_{\mathbb{N}} + I_{\mathbb{N}}: \mathbb{N} \to \mathbb{N}$ defined as $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = I_{\mathbb{N}}(x) + I_{\mathbb{N}}(x) = x + x = 2x$ is not onto.

Concept: Onto (Surjective) Functions — A function $f: A \to B$ is onto if every element in the codomain $B$ has a preimage in the domain $A$.

Reasoning:

1. $I_{\mathbb{N}}(x) = x$ maps $\mathbb{N}$ to $\mathbb{N}$. For any $y \in \mathbb{N}$, choose $x = y$; then $I_{\mathbb{N}}(x) = y$. Hence $I_{\mathbb{N}}$ is onto.

2. Now consider $(I_{\mathbb{N}} + I_{\mathbb{N}})(x) = 2x$. The codomain is $\mathbb{N}$, but the outputs are only even natural numbers. …

The key idea here is that "onto" (surjectivity) is a property of the range of a function, not just a property of the individual functions being added. When you add two functions, the outputs combine in a way that can skip values, even if each original function covered everything.

Let’s unpack this carefully.

1. What does $I_{\mathbb{N}}$ being onto mean?

   A function $f: A \to B$ is onto if every element of $B$ is the image of at least one element of $A$. For $I_{\mathbb{N}}(x) = x$, the codomain is $\mathbb{N}$ and the range is also $\mathbb{N}$ — every natural number $n$ is hit by $x = n$. So $I_{\mathbb{N}}$ is onto. No surprises here.

2. Now define $g(x) = (I_{\mathbb{N}} + I_{\mathbb{N}})(x) = 2x$.

   This is a new function from $\mathbb{N}$ to $\mathbb{N}$. The codomain is still all natural numbers, but the outputs are only the even numbers: $2, 4, 6, 8, \dots$.

3. Why is $g$ not onto?

   Take any odd natural number, say $3$. Is there any $x \in \mathbb{N}$ such that $2x = 3$? No, because $3/2 = 1.5$ is not a natural number. In fact, no odd number appears in the range of $g$. Since the codomain $\mathbb{N}$ contains odd numbers, $g$ misses them — so it’s not onto.

Method: Proving a Function Is or Is Not Onto

Use this two-directional method: to prove onto, produce a preimage for every codomain element; to prove not onto, exhibit a single codomain element that has no preimage. The question here asks you to do both, on two related functions.

Steps

Step 1: State the codomain precisely.

Onto-ness depends entirely on the declared codomain, so pin it down first (here it is $\mathbb{N}$). The range must equal this codomain for the function to be onto.

Step 2: To show onto — solve $f(x) = y$ for a general $y$.

Take an arbitrary $y$ in the codomain and construct an $x$ in the domain with $f(x) = y$. For the identity $I_{\mathbb{N}}(x)=x$, simply take $x = y$; a preimage always exists, so it is onto.

Step 3: To show not onto — find one unreachable target. …

Common Mistakes

Mistake 1: Assuming the sum of onto functions is onto.

Why it's wrong: $I_{\mathbb{N}}$ is onto, yet $I_{\mathbb{N}}+I_{\mathbb{N}}$ (i.e. $2x$) is not — its range is only the even naturals. Surjectivity is not preserved under addition. Correct approach: test $f+g$ directly against its codomain rather than inferring it from the summands.

Mistake 2: Ignoring the codomain when deciding onto-ness.

Why it's wrong: $2x$ is onto as a map onto the even numbers, but here the codomain is all of $\mathbb{N}$, which includes odd numbers it never reaches. Correct approach: always compare the range with the declared codomain, which is $\mathbb{N}$. …