(A) f is injective
(B) f is into
(C) f is surjective
(D) f is invertible

Choose the correct answer from the options given below:

Correct Answer: The correct option is (A), (C), and (D) only.

Analysis of the Image:
From the provided image, the function $f:\mathbb{N}\to \mathbb{N}$ is defined as:
$$f\left(n\right)=\{\begin{array}{ll}n-1,& \text{if}n\text{is even}\\ n+1,& \text{if}n\text{is odd}\end{array}$$ where $\mathbb{N}=\{1,2,3,4,...\}$ represents the set of natural numbers.

Let us evaluate the values of the function for the first few natural numbers:
For $n=1$ (odd): $f\left(1\right)=1+1=2$
For $n=2$ (even): $f\left(2\right)=2-1=1$
For $n=3$ (odd): $f\left(3\right)=3+1=4$
For $n=4$ (even): $f\left(4\right)=4-1=3$
In general, we see that the function swaps adjacent pairs of natural numbers: $(1\leftrightarrow 2)$, $(3\leftrightarrow 4)$, and so on.

Step 1: Check if $f$ is Injective (One-to-One) - Statement (A)
Let $f\left(n_1\right)=f\left(n_2\right)$.
- If $n_1$ is even and $n_2$ is odd, then $f\left(n_1\right)=n_1-1$ (which is odd) and $f\left(n_2\right)=n_2+1$ (which is even). Since an odd number cannot equal an even number, $n_1$ and $n_2$ must either both be even or both be odd.
- If both $n_1$ and $n_2$ are even:
$n_1-1=n_2-1\Rightarrow n_1=n_2$
- If both $n_1$ and $n_2$ are odd:
$n_1+1=n_2+1\Rightarrow n_1=n_2$
Thus, $f\left(n_1\right)=f\left(n_2\right)\Rightarrow n_1=n_2$ in all cases. This proves that $f$ is injective (one-to-one). Thus, Statement (A) is correct.

Step 2: Check if $f$ is Surjective (Onto) - Statement (C)
For any element $y$ in the codomain $\mathbb{N}$:
- If $y$ is even, then $y-1$ is an odd natural number. The function maps this pre-image as:
$f(y-1)=(y-1)+1=y$
- If $y$ is odd, then $y+1$ is an even natural number. The function maps this pre-image as:
$f(y+1)=(y+1)-1=y$
Since every natural number in the codomain has a pre-image in the domain, the range of the function is equal to the codomain $\mathbb{N}$. Therefore, $f$ is surjective (onto). Thus, Statement (C) is correct.
Since $f$ is surjective (onto), it cannot be an "into" function (which by definition is a function whose range is a proper subset of the codomain). Thus, Statement (B) is incorrect.

Step 3: Check if $f$ is Invertible - Statement (D)
A function is invertible if and only if it is bijective (both injective and surjective). Since we have verified that $f$ is both injective and surjective, it is bijective, and hence invertible. Thus, Statement (D) is correct.

Conclusion:
Statements (A), (C), and (D) are correct. Therefore, the option (A), (C), and (D) only is the correct choice.