nLab
path space object

Definition

In a category with weak equivalences and with products a path space object (often just called a path object) for an object $C$ is a factorization of the morphism $C\stackrel{\mathrm{Id}\times \mathrm{Id}}{\to }C\times C$ as

$$C\stackrel{s}{\to }{C}^I\stackrel{(d_0,d_1)}{\to }C\times C$$C \stackrel{s}{\to} C^I \stackrel{(d_0, d_1)}{\to} C \times C

such that $s$ is a weak equivalence.

Notice. Here $C^I$ is a primitive symbol. $I$ is not assumed to be an object and $C^I$ is not assumed to be an internal hom. This is standard but somewhat abusive notation. It is supposed to remind us of the “nice” situation where the path object is co-represented after all. See interval object.

If the category in question also has a notion of fibrations, such as in a category of fibrant objects or in a model category, the morphism $C^I\stackrel{(d_0,d_1)}{\to }C\times C$ in the definition of a path object is required to be a fibration.

Path space objects are in particular guaranteed to exist in any model category.

Related notions

Right homotopies

Path objects are used to define a notion of right homotopy between morphisms in a category. Thus they capture aspects of higher category theory in a 1-categorical context.

Loop space objects

From a path space object may be derived loop space objects.

Discussion

Originally the remark on abusive notation was missing and Toby asked:

What is $I$ here? Is it something that any category with weak equivalences must have (although I don't see how offhand), or is part of the data of the path object? (Indeed, it seems to be the only actual object in that data!) And then what are $d_0$ and $d_1$ exactly; maps $C^I\to C$, or maps involving $I$ itself whose product induces maps $C^I\to C$?

Urs: I hope the remark above now clarifies this. If so, this discussion part here should be removed.

Toby: The notation still doesn't make literal sense, since $C^I$ (primitive or not) isn't a product. But I believe that you just mixed up product and pairing, so I fixed that. In other words, I interpret it that $d_0$ and $d_1$ are each morphisms from $C^I$ to $C$.