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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    $\begingroup$ Why do you think it "needn't be"? $\endgroup$ – Mariano Suárez-Álvarez Nov 5 '12 at 1:23
  • $\begingroup$ $f$ is a particular arrow. $g$ is a particular arrow. $f\circ g$ is another arrow. What is the problem? $\endgroup$ – Braindead Nov 6 '12 at 5:50

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.

  • $\begingroup$ 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. $\endgroup$ – Dan Christensen Nov 5 '12 at 2:54
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    $\begingroup$ Sure. But the point is that the data of a category tells you which one is the composite. $\endgroup$ – Zhen Lin Nov 5 '12 at 8:09
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    $\begingroup$ @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. $\endgroup$ – Braindead Nov 6 '12 at 14:56
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    $\begingroup$ @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. $\endgroup$ – Braindead Nov 6 '12 at 15:08
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    $\begingroup$ @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. $\endgroup$ – Braindead Nov 6 '12 at 18:11

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.


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