# Composition of Functions, an Exercise

### Solution

Since $g\circ f\,$ is a bijection, $f\,$ is bound to be an injection.;an injection.;a surjection.;a bijection Indeed, $f(x_1)=f(x_2)\,$ $x_1\ne x_2,\,$ would imply $g(f(x_1))=g(f(x_2))\,$ in contradiction with $g\circ f\,$ being a bijection.

Since $f\,$ is an injection but not a surjection, |X|\lt |Y|.;$|X|\lt |Y|.$;$|X|=|Y|.$;$|X|\gt |Y|.$ This directly implies that $g\,$ is not an injection;an injection;a surjection;a bijection and $f,\,$ indeed, is not a surjection.;an injection.;a surjection.;a bijection.

It's given that |X|=3.;$|X|=3.$;$|X|=4.$;$|X|=5.$ Let's then try to find $Y,\,$ with |Y|=4.;$|Y|=2.$;$|Y|=3.$;$|Y|=4.$ So assume $X=\{a,b,c\}\,$ and $Y=\{u,v,w,x\}.$ Define $f\,$ with $f(a)=v,\,$ $f(b)=w,\,$ $f(c)=u.\,$ Define $g\,$ with $g(u)=g(x)=a,\,$ $g(v)=b,\,$ $g(w)=c.$

We then have g(f(a))=b,;$g(f(a))=a,$;$g(f(a))=b,$;$g(f(a)),$=c g(f(b))=c,;$g(f(b))=a,$;$g(f(b))=b,$;$g(f(b))=c,$ g(f(c))=a.;$g(f(c))=a.$;$g(f(c))=b.$;$g(f(c))=c.$ So that $g\circ f\,$ is not an identity. Thus all the conditions of the problem are fulfilled.

### Acknowledgment

This problem has been given as a homework assignment for CIS160 at UPenn, where my young son is a freshman (2016-2017).

### Functions

- Functional Notations and Terminology
- Examples of Functions
- Addition of Functions
- Multiplication of Functions
- Cantor Set and Function
- Limit and Continuous Functions
- Sine And Cosine Are Continuous Functions
- Composition of Functions, an Exercise

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny

68957325