
MHB Master
#1
March 25th, 2020,
14:12
Hey!!
For the functions $f:A\rightarrow B$, $g:B\rightarrow A$ and $h:B\rightarrow A$ it holds that $(g\circ f)(x)=x, \ \forall x\in A$ and $(f\circ h)(x)=x, \ \forall x\in B$. Show that it holds that $g\equiv h$.
I don't really have an idea how to show that.
Let $x\in A$. Then $(g\circ f)(x)=x \Rightarrow g(f(x))=x$. Then $f(x)\in B$.
We have that $(f\circ h)(x)=x, \ \forall x\in B$. Therefore $f(x)=(f\circ h)(f(x))$.
Is this the correct way to start?

March 25th, 2020 14:12
# ADS
Circuit advertisement

MHB Seeker
#2
March 25th, 2020,
14:33
Hey mathmari!!
How about a proof by contradiction?
That is, suppose $g\not\equiv h$, then there must be an $y\in B$ such that $g(y)\ne h(y)$, mustn't it?
Suppose we let $x_1 = g(y)$ and $x_2=h(y)$.
Then $x_1,x_2\in A$ and $x_1\ne x_2$.
Can we find a contradiction?

MHB Master
#3
March 25th, 2020,
14:55
Thread Author
Originally Posted by
Klaas van Aarsen
How about a proof by contradiction?
That is, suppose $g\not\equiv h$, then there must be an $y\in B$ such that $g(y)\ne h(y)$, mustn't it?
Suppose we let $x_1 = g(y)$ and $x_2=h(y)$.
Then $x_1,x_2\in A$ and $x_1\ne x_2$.
Can we find a contradiction?
Let $x_1,x_2\in A$ with $x_1\ne x_2$.
We have that $(g\circ f)(x_1)=x_1 \Rightarrow g(f(x_1))=x_1$ and $(g\circ f)(x_2)=x_2 \Rightarrow g(f(x_2))=x_2$. Since $x_1\neq x_2$ it must be $f(x_1)\neq f(x_2)$.
We have that $f(x_1), f(x_2)\in B$. It holds that $x_1=h(y_1)$ and $x_2=h(y_2)$, for some $y_1, y_2\in B$. Since $f(x_1)\neq f(x_2)$, i.e. $f(h(y_1))\neq f(h(y_2))$, it follows that $y_1\neq y_2$.
Is this correct so far? How could we continue? I got stuck right now.

MHB Seeker
#4
March 25th, 2020,
15:06
Originally Posted by
mathmari
Let $x_1,x_2\in A$ with $x_1\ne x_2$.
We have that $(g\circ f)(x_1)=x_1 \Rightarrow g(f(x_1))=x_1$ and $(g\circ f)(x_2)=x_2 \Rightarrow g(f(x_2))=x_2$. Since $x_1\neq x_2$ it must be $f(x_1)\neq f(x_2)$.
We have that $f(x_1), f(x_2)\in B$. It holds that $x_1=h(y_1)$ and $x_2=h(y_2)$, for some $y_1, y_2\in B$. Since $f(x_1)\neq f(x_2)$, i.e. $f(h(y_1))\neq f(h(y_2))$, it follows that $y_1\neq y_2$.
Is this correct so far? How could we continue? I got stuck right now.
Don't we have $x_2=h(y)$ so that $f(x_2)=f(h(y))=y$?
And therefore $g(f(x_2))=g(y)=x_1$?

MHB Master
#5
March 25th, 2020,
15:26
Thread Author
Originally Posted by
Klaas van Aarsen
Don't we have $x_2=h(y)$ so that $f(x_2)=f(h(y))=y$?
And therefore $g(f(x_2))=g(y)=x_1$?
Ahh yes! And since $g(f(x_2))=x_2$ it follows that $x_2=x_1$, which is a contradiction.
Thank you so much!!