1

I want example for some functions $f,g:X\to X$, where $X$ is an arbitrary set, such that $g(f(x))=Id_{X}(x)$ and $f$ and $g$ either not bijective (this means $f$ is bijective, $g$ is not bijective is OK. $f$ and $g$ both are bijective is No. $f$ and $g$ neither are bijective is OK.)

I think these two functions exist. but I can't find examples $f, \, g$ please find example

I am Korean so I can't use English well. sorry.

user773458
  • 1,353
Agnesi Maria Huangong
  • 29

2 Answers2

2

Consider $X = \mathbb{N}$. We define $f:X\to X$, $f(x)=2x$ and $g:X\to X$, $g(x) = \frac{x}{2}$ if $x$ is even, and $g(x) = \frac{x+1}{2}$ if $x$ is odd. We have that $g(f(x)) = g(2x) = \frac{2x}{2} = x$ for all $x \in X$, that is, $g(f(x)) = Id(x)$. But $f$ is not a surjective function, and $g$ is not an injective function. Thus, $f$ and $g$ are not bijective functions.

ZAF
  • 2,618
  • 7
  • 13
0

There is simple counterexample. Define $f,g:[0,1] \rightarrow [0,1]$ by $f(x)=\frac{x}{2}$ $(\forall x \in [0,1])$, $g(x)=2x$ $\Big(\forall x \in [0,\frac{1}{2}]\Big)$, and $g(x)=0$ $\Big(\forall x \in (\frac{1}{2},1]\Big)$. Then, $g$ $\circ$$f$ is identity map on $[0,1]$ where both $f,g$ are not bijective.

Note that $g$ $\circ$$f=id_X$ implies the injectiveness of $f$ and the surjectiveness of $g$, where $f$ is a map from a set $X$ into a set $Y$ and $g$ is a map from $Y$ into $X$. But $f,g$ may not be surjective and injective (resp).

youngchan Kim
  • 21