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From a category $\mathcal{C}$ we can construct its arrow category $\text{Ar}(\mathcal{C})$, where objects are morphisms and arrows are commutative squares.

But what happens with arrow composition? This does not correspond directly to a monoidal category structure since not every two arrows can be composed to give rise to a third one.

  • Where does the composition on arrows enters on the structure (besides the composition of morphisms between arrows themselves, that are pair of arrows)?
  • Can the composition be expressed inside this category?

Thanks!

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The arrow category is equipped with two functors $s, t : \text{Ar}(C) \to C$ giving the source and target of an arrow. Composition is a functor

$$\text{Ar}(C) \times_{\text{Ob}(C)} \text{Ar}(C) \to \text{Ar}(C)$$

where the LHS is a (2-)pullback, with one of the maps $\text{Ar}(C) \to \text{Ob}(C)$ being source and one being target. This expresses precisely that only morphisms with compatible sources and targets can be composed.

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    $\begingroup$ Do you mind if I ask how can $Ar(C) \times_{Ob(C)} Ar(C)$ be a $2$-pullback? $\endgroup$ – Giorgio Mossa Jul 14 '16 at 11:30
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One possible way to "see" composition of arrows of the original category $\mathcal{C}$ in $\operatorname{Ar}(\mathcal{C})$ is the following:

Suppose $f:x\to y$ and $g:y\to z$, let $1_x,1_y,1_z\in\operatorname{Ar}(\mathcal{C})$ be the identity morphisms of $x,y,z$ respectively, then $f$ gives a morphism $1_f:1_x\to1_y$ in the obvious way, and similarly for $g$. Then $$1_{g\circ f} = 1_g\circ1_f.$$

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It works just like any other category where arrows (here, say $(u,v): f -> g$ and $(u',v') : g -> h$) compose only if the codomain of the first (say $g:A -> B$) is the domain of the first.

This entails that $u$ composes with $u'$ and $v$ with $v'$ (hence $(u,v)$ with $(u',v')$) as they share $A$ and $B$, the extremities of $g$

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