MHB Marcus 's question at Yahoo Answers (Bijectivity on finite and infinite sets)

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For a finite set A, if the composition of two functions f and g is a bijection, then both f and g must also be bijections. This is demonstrated through the injectivity of g leading to its bijectivity, and subsequently showing that f is also bijective. In contrast, for an infinite set A, such as the integers, it is possible for f and g to compose into a bijection while neither function is bijective individually. An example provided illustrates this with specific functions f and g, where their composition results in the identity function, confirming the distinction between finite and infinite cases. The discussion emphasizes the critical difference in behavior of bijections in finite versus infinite sets.
Fernando Revilla
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Here is the question:

Supposed that A is a finite set, f: A --> A and g: A --> A. Supposed in addition that f o g: A --> A is a bijection. Prove that f and g are both bijections.
Give an explicit example to show that the conclusions of the previous problem is false if A is an infinite set. In other words, if A is an infinite set, f: A --> A and g: A --> A are functions and f o g: A --> A is a bijection it is not necessarily the case that f and g are both bijections.

Here is a link to the question:

Abstract math question: bijectivity on finite and infinite sets? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Marcus,

For a finite set $A$, any map $h:A\to A$ is injective iff it is surjective iff it is bijective. Suppose $f\circ g$ is a bijecion. Let us see that $g$ is injective $$g(x)=g(y)\Rightarrow f[g(x)]=f[g(y)]=(f\circ g)(x)=(f\circ g)(y)\Rightarrow x=y$$ That is, $g$ is injective which implies ($A$ finite) that $g$ is a bijection. Now, $$f=f\circ (g\circ g^{-1})=(f\circ g)\circ g^{-1}$$ and the composition of bijections is a bijection.

For $A$ infinite, consider $A=\mathbb{Z}$, and let

$$f:A\to A\;,\quad f(x)=\left\lfloor\frac{x}{2}\right\rfloor\\g:A\to A,\;\quad g(x)=2x$$ Then, $$(f\circ g)(x)=f(2x)=\left\lfloor\frac{2x}{2}\right\rfloor=x$$ is the identity, so $f\circ g$ is a bijection, but $f$ is not injective, for example $f(0)=f(1)=0$.

If you have further questions, you can post them in the http://www.mathhelpboards.com/f15/ section.
 
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