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In category theory, the expression $f\circ g=h$ ($\circ$ being the binary composition "function" on the class of morphisms) suggests that the morphism $h$ is unique when it clearly needn't be. Do we not have to define an equivalence relation on the set of morphisms using the $hom$ class?

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    Why do you think it "needn't be"?2012-11-05
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    $f$ is a particular arrow. $g$ is a particular arrow. $f\circ g$ is another arrow. What is the problem?2012-11-06

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Do you have some sort of counterexample or heuristic reasoning that leads you to believe that $h$ is not unique? Given a concrete category, the binary composition of morphisms becomes the binary composition of functions. Since a function takes one input and relates it to exactly one output, applying $g$ to the input $x$ gives the unique output $g(x)$, and the composition $f \circ g$ applied to $x$ gives $f(g(x))$, which is the unique output of $f$ applied to $g(x)$. Since this is the case for all objects $x$ in the concrete category in question, $f \circ g$ has been uniquely determined insofar as I can see.

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    I understand that that there may be any number of distinct morphisms (arrows) from object $X$ to object $Y$. If this is the case, then, if we have a morphism $f$ from object $A$ to object $B$, and a morphism $g$ from object $B$ to object $C$, and, say, two distinct morphisms $h1$ and $h2$ from $A$ to $C$, then either one of $h1$ and $h2$ could be seen as the composition of $f$ followed by $g$ if $h1$ and $h2$ are not seen as equivalent.2012-11-05
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    Sure. But the point is that the data of a category tells you _which one_ is the composite.2012-11-05
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    Aren't $h1$ and $h2$ *both* the composition of $f$ followed by $g$? Perhaps it's not the best definition -- I don't know -- but at the one given at Wiki, there is no such suggestion that there is an kind of selection mechanism (e.g. the use of AC) http://en.wikipedia.org/wiki/Category_theory#Categories.2C_objects.2C_and_morphisms2012-11-05
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    Or do we just accept the fact that $g\circ f$ is not *the only* composition, but only *a possible* composition?2012-11-05
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    @Dan - I honestly don't understand why you are confused about this. You have an arrow $f: A\to B$ and $g: B\to C$. Then, $g \circ f: A \to C$. If there is another arrow $h: A \to C$, this $h$ could be equal to $g\circ f$, or it could be an entirely different one. The point is that given such $f$ and $g$, there exists a unique arrow called $g\circ f$.2012-11-06
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    @Braindead It's a subtle point that seems not come up much in informal definitions and proofs. For my purposes (machine-verified formal proofs), I had to be sure that the functional notation ($g\circ f$) was justified. In another forum, my intuition was confirmed: If there are several arrows (morphisms) to choose from, the choice is arbitrary. As AK put it, "We flip a coin."2012-11-06
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    @DanChristensen - $gf$ is part of the data of the category. If you have a set of arrows from $A$ to $C$, then $gf$ is one of the element. There is no "choice" being made.2012-11-06
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    @DanChristensen - I can't help but to think that you are deeply confused. Imagine that you have a group. And there are group elements $g$ and $f$. What you are insisting is analogous to saying "There are several elements of the group to choose from [for the choice of $gf$], the choice is arbitrary." What $gf$ is part of the data of the group.2012-11-06
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    @Braindead I agree that in many categories, multiple possibilities for a composition is simply not an issue.2012-11-06
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    @DanChristensen This is my last comment here. A particular category comes with a collection of arrows with a definite composition rule. There is no choosing or coin flipping or whatever. If you change the composition rule, then you have a DIFFERENT category from the one you started out with. You are completely mistaken in thinking that "in a category, there are multiple possibilities for a composition." It simples makes no sense. It does not follow from the definition. I have no idea who told you that, but that statement is simply wrong.2012-11-06
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    @Braindead (My last comment here) The choices I allude to here are made as you are describing a structure which you intend to prove is a category (i.e. as you are defining classes of objects, morphisms, compositions, etc.).2012-11-06
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The common definition of a category demands that for every morphism $f$ there are uniquely determined domain a codomain objects. So, strictly speaking for every objects $A,B,C,D$ in a category if $Hom(A,B)\cap Hom(C,D)\ne \emptyset$ then $A=C$ and $B=D$.

Most definitions of (set enriched) categories will include either the explicit domain/codomain (large) functions or the pair-wise disjoint condition on the hom-sets as axioms (each can be proved from the other of course).

Side remark: Interestingly, in the definition of enriched categories these conditions do not appear explicitly.