I'm just beginning to dabble into some category theory (from Aluffi's Algebra) and I have some difficulty with morphisms in a slice category. Particularly, I can't see why the diagram of a morphism should commute in the ambient category. If we have the category $C_A $ and two objects

$$ f:Z\longrightarrow A $$


$$ g:Y\longrightarrow A $$

why shouldn't any morphism $Z\longrightarrow Y$ count as valid, as long as there is a morphism from both $Y$ and $Z$ into $A$, not only those $\sigma$ such that $f=g\sigma$? If the answer is obvious or I have some fundamental misunderstanding you could just say so and I'll try to think more about it.

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    $\begingroup$ Because the definition of slice categories demands this triangle to commute. You can define a different category that has the morphisms you describe, but it might not be useful. $\endgroup$ – Christoph Mar 29 '18 at 9:18

The question "why shouldn't" is a little bit difficult when talking about a definition. Just as "why shouldn't 1 be a prime number" or "why shouldn't the empty set be a vector space", the reason basically is that the definition would give something much less interesting. For instance, whenever the category has a terminal object $1$, and $A$ has a point, i.e., a morphism $1 \to A$, then the category you propose is equivalent to the category $C$ itself.

The point of the category $C_A$ is to contain something like "$A$-indexed objects of $C$". Think of this as analogous to a category of bundles (e.g. vector bundles) over a space $A$: the morphism $Z \to A$ is the projection onto the base space, and the commutativity constraint on morphisms ensures that a morphism between two bundles maps the fiber in one bundle over a point $a \in A$ into the fiber over the same point of the other bundle.

  • $\begingroup$ Thank you, I understand. When they are presented for the first time, Aluffi writes "There really is only one sensible way to assign morphisms to a category with objects as above" and invites the reader to come up with the definition himself. I guess this made me think the definition might need only elementary considerations or at least ones that require less mathematical maturity, as it is in the beginning of the book. $\endgroup$ – nek28 Mar 29 '18 at 9:26
  • $\begingroup$ @nek28, that's fair. As a rule of thumb, the definition of morphism has to "interact nicely with the object" -- everything about the object that can reasonably be "preserved" by the morphism should be. Of course, what "preserved" means can be counterintuitive without further context -- it might lead you to believe that the category of topological spaces should have open maps as morphisms, rather than continuous maps. $\endgroup$ – Mees de Vries Mar 29 '18 at 9:33

A general rule of thumb is that if you are trying to construct a category, and the objects of your category can be viewed in some reasonable way as functors, then the morphisms should probably be natural transformations.

In this situation, each object can be viewed as functors $F : \mathbf{2} \to \mathcal{C}$, where $\mathbf{2}$ is the category presented by the diagram "$0 \to 1$". The specific objects we're interested in can be identified as those functors with $F(1) = A$.

So, a very reasonable notion of morphism in the slice category should be a natural transformation between such functors; i.e. as commutative squares

$$\begin{matrix} F(0) &\xrightarrow{F(\to)}& A \\ {\ \ }\downarrow f & & {\ \ }\downarrow {g} \\ G(0) &\xrightarrow{G(\to)}& A \end{matrix}$$

The interesting question is why we should require that $g$ be the identity — that is basically because the equation $F(1) = A$ can be reasonably interpreted as an equation $\eta(1) = A$ between natural transformations as well; in the more usual notation, this equation would be written $\eta_1 = \mathrm{id}_A$.

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    $\begingroup$ In that case, OP's question can be translated as: why a transformation should be natural ? $\endgroup$ – Pece Mar 30 '18 at 4:00

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