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According to Wikipedia article on universal property,

If $U:D \rightarrow C$ is a functor and $X$ an object in $C$, then initial morphism from $X$ to $U$ is an object in category $(X\downarrow U)$ of morphisms from $X$ to $U$. In other words, it consists of a pair $(A,\Phi )$ where $A$ is an object of $D$ and $\Phi:X\rightarrow U(A)$ is a morphism in $C$.

My question is why do we even need to start with category $D$ and functor $U$? Why can't we just have the universal property with just category $C$ and it's Arrow category, since we already have everything we need in category $C$?

Here is an example to make my question more clear:

Suppose we have category $C$ with following morphisms:

$f_1: x \rightarrow U(z)$
$f_2: x \rightarrow U(b)$
$f_3: x \rightarrow U(a)$
$U(g): U(a) \rightarrow U(b)$
$U(h): U(a) \rightarrow U(z)$

Then we will have the following as the Arrow category of $C$:

objects:

$<x,f_3,U(a)>$
$<x,f_2,U(b)>$
$<x,f_1,U(z)>$

morphisms:

$U(g): <x,f_3,U(a)> \rightarrow <x,f_2,U(b)>$
$U(h):<x,f_3,U(a)> \rightarrow <x,f_1,U(z)>$

And it can be seen that $<x,f_3,U(a)>$ is an initial object in Arrow category, hence making $f_3$ an initial morphism.

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    $\begingroup$ Assuming the construction you've just suggested works, you've engineered a very particular case where the initial object of $(X\downarrow U)$ and $C^{\to}$ coincide. But just because there's an instance where it coincides does not mean it will generally. Initial objects of $(X\downarrow U)$ for arbitrary $C$ and $U$ are not usually very special in the arrow category of $C$. $\endgroup$ Nov 2, 2019 at 17:36

1 Answer 1

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There are a least a couple problems with your construction.

First, your construction uses $U$ and the objects $a$, $b$ and $z$. Where do they come from? If $U$ is still a functor and $a$, $b$ and $z$ are objects of $\mathcal{D}$, then we're back where we started with regards to the original definition. You may just be treating $U(a)$ and the rest as formal symbols, which is fine, but that means that this has nothing to do with the original construction.

Second, you don't have a complete list of the objects of the arrow category of $\mathcal{C}$. You said we morphisms $U(g) : U(a) \to U(b)$ and $U(h) : U(a) \to U(z)$, which means there are (at least) more objects in the arrow category. Also, you've neglected to mention the four identity arrows for $x$, $U(a)$, $U(b)$ and $U(z)$.

Moreover, there are a few compositions missing, though you may be implicitly saying that $U(g) \circ f_3 = f_2$ and $U(h) \circ f_3 = f_1$. We have to assume that this is true for $U(g)$ (really $\langle id_x, U(g) \rangle$) and $U(h)$ to be morphisms in the arrow category.

Now including the missing objects, $\langle x, f_3, U(a) \rangle$ is no longer initial: there's no morphism from $\langle x, f_3, U(a) \rangle$ to $\langle x, id_x, x \rangle$ since there's no morphism $U(a) \to x$ in $\mathcal{C}$. Instead, $\langle x, id_x, x \rangle$ is initial (assuming the necessary diagrams commute).


The answer to your question, though, is yes. Universal properties can be defined using a single category (no arrow category needed). In fact, initial objects encompass all or almost all universal properties. It's all just a matter of changing which category you want the initial object to be in. The construction with the functor $U$, however, is simply a useful special case (for the category $(X \downarrow U)$) that applies to many situations. For example, limits can be defined using that construction.

It's just helpful to work out the details for special cases so you get a good general idea of what's possible. "Initial objects" be themselves don't seem very interesting, but using some special categories derived from other data (in this case, the functor $U: \mathcal{D} \to \mathcal{C}$), you get something much more interesting.

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