# Thread: problem involving composite functions

1. ## problem involving composite functions

Hi

Here is a problem I am trying.

Suppose $A\neq \varnothing$ and

$f:\;A\longrightarrow A$ and

$\forall g[(g:A\longrightarrow A)\Rightarrow (f\circ g = f)]\cdots (1)$

I have to prove that f is a constant function.
I will give the outline of what I did.
Since $\because A\neq \varnothing \Rightarrow \exists a \in A$

$\therefore a \in A$

Now consider the set

$g=\{(x,a)\vert \; x\in A \; \}$

$\because g\subseteq A\times A$

g is a relation from A to A. Then I proved that g is also a function by proving
that

$\forall x \in A \exists ! b \in A ((x,b) \in g)$

next , I proved that g is a constant function by proving that

$\exists b \in A \forall x\in A (g(x)=b)$

and finally I used the given (1) for g , which implies that

$f\circ g =f$

To prove that f is a constant function, I have to prove

$\exists b\in A \forall x\in A (f(x)=b)$

since $a\in A$ , I can associate some $c\in A$ such that

$f(a)=c$

so Let $b=c=f(a)$

so the goal now is

$\forall x\in A (f(x)=b)$

Let x be arbitrary , $\therefore x\in A$

$f(x)=f(g(x))=f(a)=b$

since x is arbitrary

$\therefore \exists b\in A \forall x\in A(f(x)=b)$

so f is a constant function..................

is the proof too detailed ? since this is from Velleman's "how to prove it" , I think,
author expects me to use all the logical machinery that I can use.

correct ?

2. ## Re: problem involving composite functions

It's correct. Just as a matter of style though, it could be much more terse:

Since A is non-empty, let a be in A.
Let g = {<x a> | x is in A}.
g is a function from A into A.
Suppose x is in A.
So f(x) = f(g(x)) = f(a).
So f is a constant function.

3. ## Re: problem involving composite functions

thanks moeblee......