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I am playing around with equations about functions in general and have some questions.


Question 1

If I have some functions $f,g\colon X^2 \rightarrow Y$ such that $f(x,y) = g(x,c)/g(y,c)$ then can $f(x,y)$ always be written as $f(x,y) = h(x)/h(y)$ for some function $h\colon X\rightarrow Y$? Since $f(x,y)$ is not dependent on $c$, shouldn't it always be possible to write a $c$-free expression for it? How can this be proven? Is there some theorem about this or to support this?


Question 2

Say, I had some function $f \colon X^2 \rightarrow Y$ and $x,y,z \in X$. Now if I have some equation like this,

$$f(x,y) \cdot f(y,z) = f(x,z)$$

where the equation is made up entirely of expressions made up of the function. Then, is this equivalent to

$$ a \cdot b = c $$

where $a = f(x,y)$, $b = f(y,z)$, and $c = f(x,z)$ and so $a,b,c \in Y$? And therefore, is it okay to work with the equation based on the properties of $Y$? If that is true, then the elements $x,y,z$ don't play a role in the equation if they are defined to be "any elements in $X$", right?


Question 3

Say, we have some set $X$, some monoid $(Y,\cdot)$, and some function $f\colon X \rightarrow Y$.

$$f(x)\cdot f(y) = f(z) \implies f(x) = f(z) f(y)^{-1} $$

Is the above statement necessarily wrong? It should be, I think, but I am not sure.

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  • $\begingroup$ For Question 3: a monoid need not have inverses, so the conclusion may not make sense. If $Y$ is a group though, then the implication is true (so it is not necessarily wrong) $\endgroup$
    – zcn
    Oct 21, 2014 at 7:40
  • $\begingroup$ So if something is a monoid, but NOT a group - then the statement is definitely wrong, yes? $\endgroup$
    – XYZT
    Oct 21, 2014 at 13:31

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