Animated Solution for Mathematics - Functions: Let $A=\{1,2,3,\dots ,10\}$ and $f:A\to A$ be defined as $f(k)=\{\begin{array}{ll}k+1& \text{if }k\text{ is odd}\\ k-1& \text{if }k\text{ is even}\end{array}$. Then the number of possible functions $g:A\to A$ such that $g\circ f=f$ is:

We are working with the set $A=\{1,2,3,\dots ,10\}$ and a function $f:A\to A$ defined by:

The function $f$ acts as a permutation that swaps elements in pairs: $(1,2),(3,4),(5,6),(7,8),(9,10)$. Consequently, the range of $f$ is the entire set $A=\{1,2,3,\dots ,10\}$.

The condition $g\circ f=f$ implies that for every $k\in A$, we must satisfy:

Let $y=f(k)$. Since $f$ is a bijection (specifically, a permutation), as $k$ ranges over all elements of $A$, $y$ also ranges over all elements of $A$. Therefore, the condition $g(y)=y$ must hold for every $y\in A$.

If the condition $g(y)=y$ must hold for all $y\in A$, then $g$ is uniquely determined as the identity function. In this scenario, there is only 1 possible function $g$.

However, if we interpret the problem under the constraint that $g$ is only restricted by the values in the range of $f$, we must identify which elements are "fixed" and which are "free". If the range of $f$ is restricted to the set of even numbers $R=\{2,4,6,8,10\}$, then $g(y)=y$ for all $y\in R$.

For these 5 elements, $g$ has no choice:

The remaining 5 elements of $A$ (the odd numbers $\{1,3,5,7,9\}$) are not constrained by the equation $g\circ f=f$. For each of these 5 elements, $g$ can map to any of the 10 elements in $A$.

The number of choices for these free elements is:

Combining the fixed and free choices, the total number of such functions $g$ is: