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Proving a nice functional property

A function g from a set $$X$$ to itself satisﬁes $$g^{m} = g^{n}$$ for positive integers m and n with $$m > n$$. Here $$g^{n}$$ stands for g ◦ g ◦ · · · ◦ g (n times). Show that g is one-to-one if and only if g is onto. (Some of you may have seen the term “one-one function” instead of “one-to-one function”. Both mean the same.)

Note by Eddie The Head
3 years, 11 months ago

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Only If part. Assume $$g$$ is $$1-1$$. We will show that $$g^{-1}$$ exists and this will prove the claim.

Since $$g$$ is $$1-1$$, so is $$g^{j}$$, for any positive integer $$j$$. Now, we have for all $$x\in X$$, and the given $$m$$ and $$n$$, $$g^m(x)=g^n(x)$$, i.e. $$g^{n}\odot g^{m-n}(x)= g^{n}(x)$$. From the $$1-1$$ property of $$g^{n}$$, we get $$g^{m-n}(x)=x$$, for all $$x\in X$$. Hence, $$g^{-1}$$ exists and is given by $$g^{m-n-1}$$ (if $$m=n+1$$, $$g$$ is then the identity mapping). This proves the claim.

If part can be solved by similar ideas.

- 3 years, 11 months ago

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Nice!

- 3 years, 11 months ago

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