# Resemblance between product and homotopy

The notion of product $X\times X$ for an object $X$ of a category $C$ resembles the notion of homotopy between two continuous functions. Indeed the relevant diagrams look the same:

\begin{array}{ccccccc} &&X&&&&X\\ &\nearrow&\uparrow&&&\nearrow&\uparrow\\ Y&\dashrightarrow&X\times X&&Y&\dashrightarrow&X^{[0,1]}\\ &\searrow&\downarrow&&&\searrow&\downarrow\\ &&X&&&&X \end{array}

On the left: $X$ and $Y$ are objects of a category, the vertical arrows should be thought of as "projections" on each of the two factors, the diagonal arrows are any two maps and the dashed arrow is the unique map factoring both (definition of product via universal property).

On the right: $X$ and $Y$ are topological spaces, the map space $X^{[0,1]}$ carries the compact-open topology, the vertical arrows are the two evaluation maps (at 0 and at 1), the diagonal arrows are two homotopic maps and the dashed arrow is a homotopy between the two.

We can also revert the arrows and get the same resemblance, this time between the notion of coproduct and, again, homotopy (provided we substitute $X^{[0,1]}$ with $X\times [0,1]$ etcetera).

My (admittedly vague) question is then the following:

Is there something interesting behind this resemblance, perhaps in a more abstract setting?

I recognise that the two situations are considerably different: in the left diagram, if the product object exists then any two maps (the diagonal arrows) admit a unique dashed arrow. In the right diagram, instead, only some pairs of maps admit a dashed arrow and this arrow is usually highly non-unique. However, the striking resemblance of the two diagrams encouraged me to ask the question. Perhaps an answer will involve the adjunctions between the product or coproduct functor and the diagonal functor?

• There is nothing interesting to say here, except perhaps to notice that $X \times X \cong X^2$. Oct 28 '14 at 12:28
• Perhaps ncatlab.org/nlab/show/path+space+object applied to model categories? If we endow a category with the trivial model structure, then $X^I$ is just $X$ and not $X \times X$. Is there another model structure in which $X^I$ becomes $X \times X$? Oct 28 '14 at 21:47

I see your vague question and raise you a vague answer. The punchline is that you should think of $X^{[0, 1]}$ as being analogous, not to $X \times X$, but to $X$.

If $f : Y \to X$ and $g : Y \to X$ are two maps, consider the following two diagrams. First, consider the diagram

$$Y \xrightarrow{(f, g)} X \times X \xleftarrow{\Delta} X$$

where $\Delta$ is the diagonal map. Then $(f, g)$ lifts to a map $Y \to X$ if and only if $f = g$; moreover, if such a lift exists, it is necessarily unique, so the space of such lifts is either empty or a point (it's a truth value).

Second, consider the the diagram

$$Y \xrightarrow{(f, g)} X \times X \xleftarrow{(p(0), p(1))} X^{[0, 1]}$$

where $p : [0, 1] \to X$ is a continuous path, so an element of $X^{[0, 1]}$, and $(p(0), p(1))$ is the map sending a path to its endpoints. This diagram can be thought of as a "resolution" of the previous diagram. Now a lift of the map $(f, g)$ to a map $Y \to X^{[0, 1]}$ is precisely a homotopy between $f$ and $g$; moreover, the space of such lifts is precisely the space of such homotopies.

So, by resolving the diagonal map $\Delta$ using the endpoint map $(p(0), p(1))$, we get that the "derived functor" of asking whether two maps are equal is asking for the space of homotopies between two maps.

Playing this game in more generality leads to explicit constructions of various homotopy limits and, dually, of various homotopy colimits in spaces; in particular you aren't far from here to a construction of the double mapping cocylinder or double mapping path space, which is an explicit construction of the homotopy pullback in spaces.