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Let $A=\left\{1,2,\cdots,10\right\}$

Let $f,g:A\to A$. Consider the equivalence relation

$$ fRg \iff \exists h:A\to A. f=h\circ g$$ where $h$ is invertible.

Now, let $g(x)=5$:

Why is $\left| \left\{ f\in A\to A : fRg \right\} \right| = 1$?

Update:
If $f(x) = 5$ then it's true that $f = h\circ g$ if $h$ is the identity function.
Now, Why any other function won't work?

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  • $\begingroup$ That is not an equivalence relation. Call it relation anyways. And I think that you swished the order of the relation, so it should be fRg inside the set, maybe? And try to find any f in the set. $\endgroup$
    – PenasRaul
    Sep 5, 2014 at 15:13
  • $\begingroup$ Sorry, I added another condition: $h$ is invertible. $\endgroup$
    – Aharon
    Sep 5, 2014 at 15:16
  • $\begingroup$ I corrected it to $fRg$ but anyway, it's symmetric (because $R$ is an equivalence relation). $\endgroup$
    – Aharon
    Sep 5, 2014 at 15:18
  • $\begingroup$ So I think that the answer is ten, any constant function would suit the requirements $\endgroup$
    – PenasRaul
    Sep 5, 2014 at 15:18
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    $\begingroup$ That would held 10 different functions, one for each possible value of $h$. :/ $\endgroup$
    – PenasRaul
    Sep 5, 2014 at 15:24

1 Answer 1

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I hope that you are defining the relation by $f\sim g \Rightarrow \exists h | f = g \circ h$.

If so, take a $f = g \circ h$, then $im f \subseteq im g = \{ 5 \} $ so clearly $f$ would be the constant function with value $5$, hence $f=g$.

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  • $\begingroup$ No, $f~g \iff f = h\circ g$ $\endgroup$
    – Aharon
    Sep 5, 2014 at 15:29
  • $\begingroup$ So you saw Mauro's answer, it's not 1, it's 10. $\endgroup$
    – PenasRaul
    Sep 5, 2014 at 15:30

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