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 $C$:




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

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.

  • 1
    $\begingroup$ 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 $C$. $\endgroup$ – Malice Vidrine Nov 2 '19 at 17:36

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.