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I have come across the following formula in Graph Theory and am trying to decipher what it means.

$f(x\circ y)=f(x)\circ f(y)$

It is fairly simple but I have received a number of conflicting definitions of the "$\circ$" symbol which is making it hard to interpret. In some cases people where using this to represent contraction, some to represent relation, etc.

The source can be found here in Definition $1.3$.

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    $\begingroup$ It looks like the group operation $\endgroup$
    – Henry
    Mar 16 '20 at 20:44
  • $\begingroup$ Please make us clear where exactly is the trouble? Many pages? in which page? $\endgroup$
    – Mikasa
    Mar 16 '20 at 20:44
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    $\begingroup$ Here $\circ$ is any group operation - any operation satisfying Def 1.1. $\endgroup$ Mar 16 '20 at 20:44
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    $\begingroup$ At this point, it has nothing to do with graphs yet, just groups. $\endgroup$
    – saulspatz
    Mar 16 '20 at 20:50
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It seems that $\circ$ is the group operation here.

In the homomorphism $f:G \to G^\prime$ between two groups, with $f(x\circ y) =f(x)\circ f(y)$,

  • the first $\circ$ is the group operation for $G$ and
  • the second $\circ$ is the group operation for $G^\prime$
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You are dealing with mappings between groups as defined in Def. $1.1$ in the linked document. A group is a set $S$ equipped with an inner operation $\circ\colon S\times S\to S, (a,b)\mapsto a\circ b$ satisfying some axioms (the ones given in your document). One may write $(S,\circ)$ denoting which set we are dealing with and in particular which operation is used. Now consider a different set $T$ with an inner operation $\star$ also satisfying the given axioms, making $(T,\star)$ a group aswell.

One can immediately consider usual set-functions $f\colon S\to T$ assigning to each element $s\in S$ and element $t:=f(s)\in T$. What is defined in Def. $1.3$ is a so-called group homomorphism. This is nothing but a set-function between $S$ and $T$ preserving the structure given by the inner operations $\circ$ and $\star$, respectively. This presevation is given by the axiom $f(a\circ b)=f(a)\star f(b)$ $\forall a,b\in S$; it is equal to first computing $a\circ b$ and then applying the set-function $f$ or first applying the set-function $f$ to both elements and the computing $f(a)\star f(b)$. To put it different, it is not important which way we choose to go in the following diagram(s)

$$\require{AMScd} \begin{CD} S\times S @>{f\times f}>> T\times T\\ @V{\circ}VV @VV{\star}V \\ S @>>{f}> T \end{CD}~~\implies~~ \begin{CD} (a,b) @>{f\times f}>> (f(a),f(b))\\ @V{\circ}VV @VV{\star}V \\ a\circ b @>>{f}> f(a\circ b)=f(a)\star f(b) \end{CD}$$

In the given case two groups are considered, $G$ and $G'$, where the operation is denoted by $\circ$ in both cases (this is a quite common abuse of notation, as it was already discussed yesterday for example). $f$ is a set-function $f\colon G\to G'$ such that $f(x\circ_G y)=f(x)\circ_{G'} f(y)$ (for emphasis I denoted $\circ$ accordingly, in constrast to the author's choice). Note, however, authors tend to not specify which operation is referred to leading to the axiom $f(x\circ y)=f(x)\circ f(y)$ $\forall x,y\in G$.

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    $\begingroup$ Breaking out the commutative diagrams sorta seems like overkill for helping someone who is only just learning what a group is. IMHO. $\endgroup$ Mar 16 '20 at 21:19
  • $\begingroup$ @JairTaylor It might help to visualize things, but yeah, arguably it's a complete overkill. $\endgroup$
    – mrtaurho
    Mar 16 '20 at 21:21
  • $\begingroup$ I have to admit this is a bit over my head, but it is a perfect answer and gives me a lot of areas to do more research into so exactly what I was hoping for! $\endgroup$
    – JFreeman
    Mar 17 '20 at 8:08
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    $\begingroup$ @JFreeman As soon as you stumble into new questions: ask them here on MSE! That's afterall the purpose of this site :) $\endgroup$
    – mrtaurho
    Mar 17 '20 at 13:20
  • $\begingroup$ @mrtaurho - coming back to your answer after doing more research it makes tonnes of sense and is incredibly helpful! So thank you again! $\endgroup$
    – JFreeman
    Apr 20 '20 at 23:53

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