If the function $f: \mathbb{N} \rightarrow \mathbb{N}$ is defined as $f(n)=\left\{\begin{array}{ll}n-1, & \text { if } n \text { is even } \\ n+1, & \text { if } n \text { is odd }\end{array}\right.$, then

(A) f is injective

(B) f is into

(C) f is surjective

(D) f is invertible

To understand this function conceptually, it maps even numbers to the odd number just below them, and odd numbers to the even number just above them. For example:

• $f(2) = 1$
  • $f(3) = 4$
  • $f(4) = 3$
  • $f(5) = 6$

This creates a pattern where $f$ essentially "swaps" consecutive pairs of natural numbers.

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Property (A): Is $f$ injective?

A function is injective (one-to-one) if whenever $f(a) = f(b)$, we must have $a = b$.

For our function, if $f(a) = f(b)$:

Case 1: If both $a$ and $b$ are even, then $f(a) = a-1$ and $f(b) = b-1$

So $a-1 = b-1$ implies $a = b$

Case 2: If both $a$ and $b$ are odd, then $f(a) = a+1$ and $f(b) = b+1$

So $a+1 = b+1$ implies $a = b$

Case 3: If $a$ is even and $b$ is odd, then $f(a) = a-1$ and $f(b) = b+1$

Here, $a-1$ is odd and $b+1$ is even, so they cannot be equal

Therefore, $f$ is injective.

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Property (B): Is $f$ into?

A function is "into" if it maps to a proper subset of the codomain, meaning some elements in the codomain have no preimage.

Let's analyze what values $f$ can produce:

• For even inputs $n = 2k$: $f(2k) = 2k-1$ (all odd numbers $\geq 1$)
  • For odd inputs $n = 2k+1$: $f(2k+1) = 2k+2$ (all even numbers $\geq 2$)

Together, these cover all natural numbers, so $f$ maps onto the entire codomain $\mathbb{N}$.

Since $f$ doesn't map to just a proper subset, it is not "into".

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Property (C): Is $f$ surjective?

A function is surjective (onto) if every element in the codomain has at least one preimage.

For any natural number $y$:

If $y$ is odd: We can find its preimage as $2k$ where $y = 2k-1$, so $k = \dfrac{y+1}{2}$

If $y$ is even: We can find its preimage as $2k-1$ where $y = 2k$, so $k = \dfrac{y}{2}$

Since every element in the codomain $\mathbb{N}$ has a preimage, $f$ is surjective.

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Property (D): Is $f$ invertible?

A function is invertible if and only if it's bijective (both injective and surjective).

We've already shown that $f$ is both injective and surjective, therefore $f$ is invertible.

The inverse function would be:

$f^{-1}(n) = \begin{cases} n+1, & \text{if } n \text{ is odd} \\ n-1, & \text{if } n \text{ is even} \end{cases}$

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Comparing all properties:

• (A) Injective: Yes
  • (B) Into: No
  • (C) Surjective: Yes
  • (D) Invertible: Yes

The answer is: (A), (C) and (D) only