Take $G = \{f_1,f_2,f_3,f_4,f_5,f_6\}$ be a groupoid with the operation of function composition ($\circ$), and let $F = \{f_1,f_2,f_3\}$ with $\circ$ also be a groupoid. I would like to show that there is no homomorphism between $(G,\circ)$ and $(F,\circ)$ Given that the previous functions are defined as such:

Each mapping maps values from $\mathbb{R} \setminus \{0,1\}$ into $\mathbb{R}$: $$f_1 : x \rightarrow x,f_2: x \rightarrow \frac{1}{1-x}, f_3 : x \rightarrow \frac{x-1}{x},$$ $$f_4: x \rightarrow \frac{1}{x}, f_5: x \rightarrow \frac{x}{x-1}, f_6: x \rightarrow 1-x.$$

I've been having some difficulties with this question since I am used to dealing primarily with groups, where the operation is defined for all elements of the group. Assume for the sake of contradiction such a homomorphism $h: G \rightarrow F$ exists. We see that $\circ$ generally has issues relating to its partial-ness when we use functions that have the potential to output $0$ and $1$, Although I am not entirely sure how this factors into the behavior of this homomorphism. Is it possible that I should be using some combinatorial arguments about the potential mapping $h$ can take on? Any suggestions would be appreciated.


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