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  1. MHB Master
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    #1
    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?

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  3. MHB Seeker
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    Klaas van Aarsen's Avatar
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    #2
    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?

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    #3 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    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.

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    #4
    Quote Originally Posted by mathmari View Post
    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 $g(f(x_2))=g(y)=x_1$?

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    #5 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    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!!

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