I'm working on a question in an intro algebraic topology book:

Verify that the "arrow category" is a category. That is, if $\mathcal{C}$ is a category, then show that the following construction gives a category $\mathcal{M}$: Let $\operatorname{obj} \mathcal{M}$ be the collection of arrows of $\mathcal{C}$. For $f,g\in \operatorname{obj} \mathcal{M}$, say $f: A \to B$ and $g: C \to D$, let an arrow in $\mathcal{M}$ from $f$ to $g$ be an ordered pair of arrows $(\alpha,\beta)$ (with $\alpha:A \to C$ and $\beta: B \to D$) in $\mathcal{C}$ such that $g \circ \alpha = \beta \circ f$.

Composition of arrows of $\mathcal{M}$ is given by the rule $(\alpha',\beta') \circ ( \alpha,\beta) = ( \alpha' \circ \alpha,\beta'\circ\beta)$.

My trouble is showing that the hom-sets are pairwise disjoint. As I understand it, $ \operatorname{hom} (f,g) \cap \operatorname{hom} (h,k)$ is empty if $(f,g)\neq (h,k)$.

I tried assuming the contrary just to get a feel for why this must be true: Suppose $(\alpha,\beta) \in \operatorname{hom} (f,g) \cap \operatorname{hom} (h,k)$ and $(f,g) \neq (h,k)$. If $f,g,\alpha,\beta$ have the domains and codomains given above, then we must have $h: A \to B$ and $k: C\to D$ for the composites to make sense.

The diagrams commute and hence $g \circ \alpha=\beta \circ f$ and $k \circ \alpha = \beta \circ h$. Since $(f,g) \neq (h,k)$, we have $f \neq h$ or $g \neq k$.

Then...nothing. I'm not sure that this line of attack is going to work. It seems like the assumption that $(f,g) \neq ( h,k)$ gives me nothing usable.

Perhaps I have a fundamental misunderstanding of categories or the construction of $\mathcal{M}$. If somebody could give me a nudge in the right direction I would greatly appreciate it.

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    $\begingroup$ by the way, what book is it? $\endgroup$ – magma Jan 12 '12 at 2:40

If you define a morphism of the arrow category to be an ordered pair $(\alpha,\beta)$ of a morphism from $A$ to $C$ and a morphism from $B$ to $D$ such that $g\circ \alpha=\beta\circ f$, then the homsets may not be disjoint. For example, let $\cal C$ be the category of abelian groups, $\Bbb Z$ the group of integers under addition, and $1$ the identity map on $\Bbb Z$. Then if $2: {\Bbb Z}\to{\Bbb Z}$ is the multiplication-by-2 map, $(2,2)$ is a morphism from $1$ to $1$, but $(2,2)$ is also a morphism from $2$ to $2$. The fix for this problem is to disjointify the homsets by adding information to each morphism of the arrow category to indicate unambiguously which pair of objects of the arrow category it is a map between. For example, you could take an arrow from $f$ to $g$ to be a commutative square, i.e., a 4-tuple $(f,\alpha,g,\beta)$ such that $f$, $g$, $\alpha$, and $\beta$ are all morphisms of $\cal C$, $g \circ \alpha$ and $\beta \circ f$ are defined, and $g \circ \alpha=\beta \circ f$.

  • $\begingroup$ Thank you very much, David. I'm not sure what to take away from this exercise now though. The text says that this is a category, but how can that be if the homsets aren't necessarily disjoint? $\endgroup$ – John Myers Jan 11 '12 at 3:40
  • $\begingroup$ Fix the small bug in the definition first. Then, it will be a category. $\endgroup$ – David Moews Jan 11 '12 at 3:54
  • $\begingroup$ I just found that planetmath.org's definition of the arrow category coincides with my text's: the morphisms are ordered pairs. I'm trying to reconcile the counterexample with the definitions of the arrow category and a category. Is the identification of an ordered pair with its source and target objects (the 4-tuple, the fix) somehow implicit in these definitions? Perhaps that's a silly question. $\endgroup$ – John Myers Jan 11 '12 at 4:41
  • $\begingroup$ planetmath definition uses the functor category approach, which is correct in itself, but then unfortunately goes on to say that a morphism is just a special ordered pair of morphisms, which is not correct, as i explain in my answer below $\endgroup$ – magma Jan 12 '12 at 2:14
  • $\begingroup$ @JohnM: Hom-sets need not be disjoint, because we never ever need to work with the totality of "all arrows" in a category. It is perfectly valid to define a small category to be a set of objects together with a hom-set for each pair of objects, an identity arrow for each object and (total!) composition operators for each composable pair of hom-sets, all this satisfying the obvious axioms. $\endgroup$ – Zhen Lin Jan 12 '12 at 8:14

You are right in being confused and @David was quick to point out the problem. The truth is that many (most?) textbooks of Category Theory (and of other areas of mathematics) are sometimes imprecise in their formal definitions. The reason is that their information is supposed to be "processed" by human brains which are famously rather flexible and forgiving of little imprecisions. You realise this (for example) when you start to code symbolic mathematics/logic with a computer algebra system (CAS). With a CAS you have to be very precise, so you discover all these little imprecisions present in textbooks. You were very good at spotting this one by manually doing the checks on the Hom sets. Most textbooks do not do that, they just check that you have identities and that composition makes sense.

The Arrow category $\mathcal{C}^\rightarrow$ can be described as a special comma category $\left(\text{Id}_C\downarrow \text{Id}_C\right)$ (you can see this for ex. in wikipedia: comma category). Now, morphisms in comma categories are defined "everywhere" - even in Mac Lane's Categories for the working mathematician - as special ordered pairs (g,h). For ex. in wikipedia:

We can form the comma category $(S \downarrow T)$ as follows: The objects are all triples (α,β,f) with α an object in $\mathcal{A}$, β an object in $\mathcal{B}$, and $f : S(\alpha)\rightarrow T(\beta)$ a morphism in $\mathcal{C}$. In comma category he morphisms from (α,β,f) to (α',β',f') are all pairs (g,h) where $g : \alpha \rightarrow \alpha'$ and $h : \beta \rightarrow \beta'$ are morphisms in $\mathcal A$ and $\mathcal B$ respectively, such that the following diagram commutes....

Now in category theory you are supposed to be able to recover the domain and codomain info from the name (that is symbol or formula) of a morphism. But it is pretty clear that if you are given a morphism named (g,h) with g and h as above, you are never going to recover the information about f and f'. At best you can recover the α,β,α',β' info from the g and h. So, as @David pointed out, one must name the morphisms as 4-tuples (g,h,f,f') to uniquely link them with domain and codomanin info in $\mathcal{C}^\rightarrow$.

A better/alternative definition of the comma category $\mathcal{C}^\rightarrow$ is as the functor category $\mathcal{C}^2$ (functors from category 2 to $\mathcal{C}$). In this case the morphisms are the natural transformations from two functors. Also in this case most (all?) textbooks tell you that in order to define a natural transformation all you need is a special family of morphisms (in this case just 2 morphisms). But again this is imprecise. To be precise one also need to specify the source and target functors and in this case the f and f' would do it.

I hope this clarifies the situation a bit for you

  • $\begingroup$ Thank you very much for the explanation! The book is Introduction to Algebraic Topology by Rotman. I'll take a look at the problem again (armed with more knowledge) and come back if I have any more questions. $\endgroup$ – John Myers Jan 12 '12 at 7:45

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