If $g: X \to X$ such that $g^m=g^n$ for positive integers $m$ and $n$ where $m \gt n$. $g^m$ denotes $g(g(g(\cdots g(x))$
Prove that $g$ is One to one function if and only if $g$ is Onto.
First i considered $m=2$ and $n=1$. So
$$g(g(x))=g(x) \tag{1}$$
Now Since $g$ is Onto $\forall$ $y \in X$, $\exists$ $x \in X$ such that
$$g(x)=y$$
So substituting $g(x)=y$ in $(1)$ we get
$$g(y)=y$$ which is an identity function and hence $g$ is one to one.
Similarly what ever $m$ and $n$ we choose we get $g$ as an identity function.
is this correct approach?
Assumptions in question: $g^m(x) = g^n(x)$ for all $x \in X$ for some fixed $m > n$.
Answer: Now, note that $g^m(x) = g^{m-n}g^n(x) = g^n(x)$ and thus, $g^{m-n}\restriction_{g^n(X)} = id\restriction_{g^n(X)}$.
If $g$ not onto then $g^n(X) \not= X$, and so $g^{m-n}$ pushes points outside $g^n(X)$ into $g^n(X)$ but leaves things inside $g^n(X)$ unchanged, so $g^{m-n}$ is not one to one, and thus neither is $g$. That shows one direction.
On the other hand, if $g$ is onto, then $g^n(X) = X$, and thus, since $g^{m-n} \restriction_{g^n(X)} = id \restriction_{g^n(X)}$, we see that $g^{m-n} = id$. Thus, $g$ must be one to one. That proves the other direction.