Animated Solution for Mathematics - Functions: Let N be the set of natural numbers and two functions f and g be defined as $f,g:N\to N$ such that $f(n)=\{\begin{array}{ll}\frac{n+1}{2}& \text{if }n\text{ is odd}\\ \frac{n}{2}& \text{if }n\text{ is even}\end{array}$ and $g(n)=n-(-1{)}^n$. The fog is :

Welcome, fellow traveler in the realm of mathematics! Today, we are going to dissect a beautiful problem involving composite functions. We are dealing with two functions, $f$ and $g$, both mapping from the set of natural numbers $\mathbb{N}$ to itself.

Our mission is to uncover the nature of the composite function $(f\circ g)(n)=f(g(n))$. A function is simply a rule—a process that takes an input and transforms it into an output. When we compose them, we are essentially chaining these machines together.

First, let's look at the inner function, $g(n)=n-(-1{)}^n$. This function is a bit of a trickster because the term $(-1{)}^n$ indicates that the behavior depends entirely on the parity of $n$.

If $n$ is odd, then $(-1{)}^n=-1$. Substituting this into our function, we get: Since adding $1$ to an odd number results in an even number, $g$ maps odd inputs to even outputs.

If $n$ is even, then $(-1{)}^n=1$. Substituting this, we get: Since subtracting $1$ from an even number results in an odd number, $g$ maps even inputs to odd outputs. This is a perfect parity-swapping machine.

Now, we feed these results into the outer function $f(n)$. Recall that $f(n)$ is defined as:

For an odd input $n$, we know $g(n)=n+1$, which is even. We then feed this even number into $f$:

For an even input $n$, we know $g(n)=n-1$, which is odd. We then feed this odd number into $f$:

Look at what we have just derived. For odd $n$, the result is $\frac{n+1}{2}$, and for even $n$, the result is $\frac{n}{2}$.

This is exactly the definition of our original function $f(n)$. We have discovered that $(f\circ g)(n)=f(n)$. The inner function $g$ was just a shuffle, but the composite function $f\circ g$ behaves exactly like $f$ itself.

Now that we know $(f\circ g)(n)=f(n)$, testing for injectivity (one-one) and surjectivity (onto) becomes much easier. To check if it is one-one, we ask: does every unique input map to a unique output?

Let's test $n=1$ and $n=2$: Since two different inputs map to the same output, the function is not one-one. It is many-one.

To check if it is onto, we ask: can every natural number $y$ in the codomain be reached? Let's pick any $y\in \mathbb{N}$. We need to find an $n$ such that $(f\circ g)(n)=y$.

If we choose $n=2y$, which is always an even number, then: Since we can always find a pre-image for any $y$, the function is onto.

We have journeyed through the logic of parity, simplified a composite function, and tested its fundamental properties. We found that the function is onto but not one-one. This corresponds to option (4).