# Does initial morphism imply that the object in the domain of originating functor is an initial object as well?

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:

$$$$
$$$$
$$$$

morphisms:

$$U(g): \rightarrow $$
$$U(h): \rightarrow $$

And it can be seen that $$$$ is an initial object in Arrow category, hence making $$f_3$$ an initial morphism.

• 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 $C$. – 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.