By a "homotopy commutative diagram," I mean a functor $F: \mathcal{I} \to \mathrm{Ho}(\mathrm{Top})$ to the homotopy category of spaces. By a "strictification," I mean a lifting of such a functor to the category $\mathrm{Top}$ of topological spaces. I am curious about simple instances where such a "strictification" does not exist: that is, there are obstacles to making a diagram that commutes only up to homotopy strictly commute.
For example, it is known that a homotopy coherent diagram in topological spaces can always be strictified. In other words, if one imposes homotopy commutativity, but also keeps track of all the homotopies and requires that they satisfy compatibility conditions with one another, then one can strictify. More precisely, a homotopy coherent diagram $\mathcal{I} \to \mathrm{Top}$ can be described as a simplicial functor $\mathfrak{C} \mathcal{I} \to \mathcal{Top}$, where $\mathfrak{C}\mathcal{I}$ (notation of HTT) is a "thickened" version of $\mathcal{I}$ where the usual relations in $\mathcal{I}$ only hold up to coherent homotopy. Then it is a theorem of Dwyer and Kan that the projective model structures on homotopy coherent diagrams (i.e. $\mathrm{Fun}(\mathfrak{C} \mathcal{I}, \mathrm{SSet})$) and on commutative diagrams (i.e. $\mathrm{Fun}(\mathcal{I}, \mathrm{SSet})$ are Quillen equivalent under some form of Kan extension.
Part of this question is to help myself understand why homotopy coherence is more natural than homotopy commutativity. One example I have in mind is the following: let $X$ be a connected H space, and suppose $X $ is not weakly equivalent to a loop space. Then one can construct a simplicial diagram in the homotopy category given by the nerve construction applied to $X$ (as a group object in the homotopy category). This cannot be lifted to the category of spaces, because then Segal's de-looping machinery would enable us to show that $X$ was the loop space of the geometric realization of this simplicial space.
However, this example is rather large. Is there a simple, toy example (preferably involving a finite category) of a homotopy commutative diagram that cannot be strictified?
Best Answer
EDIT: Tom Goodwillie, in the comments, points out (interpreted in a quite charitable way) that there are two mistakes with the following argument. The $\pi_1$-obstruction does exist. However:
It misreads the question and assumes that there is some portion of an actual diagram which is homotopy commutative.
Even given that, it makes a mistake: it asserts that we can ignore the indeterminacy. This is wrong. For any space in the diagram, the space of maps to the terminal object $S^1$ is not simply-connected, but instead is homotopy equivalent to $S^1$. This allows you to simply erase the obstruction in $\pi_1$ by simply making different choices of homotopies.
As such, it's advisable that what's written below be demoted and I'll try to get a correct version later.
One of the classical obstructions to realizability is for cubical diagrams. Here, the category $I$ is the poset of subsets of $\{0,1,2\}$. Given such a diagram which is homotopy commutative, you can get twelve maps (one per edge in the cube) and six homotopies (one per face, which are well-defined up to "multiplication by an element in $\pi_1$") and the collection of all possible ways to compose maps, or compose maps with homotopies, gives rise to a hexagon in the space $Map(X,Z)$. Here $X$ is the image of initial object and $Z$ is the image of the terminal object in the cube. If the diagram is actually commutative then you can choose your homotopies to be trivial, and get a trivial hexagon; if it is equivalent to an honestly commutative diagram then you can choose your six homotopies on faces so that the hexagon in $Map(X,Z)$ can be filled in with a disc, or equivalently "represents the trivial element in $\pi_1$".
All scare quotes in the above are places where I'm neglecting basepoints. You have to be more careful in a real-world situation.
Here's an example where all the spaces in the diagram are contractible, except for the terminal one. This makes it easy to ignore the indeterminacy.
Let $Z$ be $S^1$, viewed as a quotient of $[0,1]$. We have six subsets of $Z$, which are the images of $$ [0,1/3], [1/3, 2/3], [2/3, 1], \{0\}, \{1/3\}, \{2/3\} $$ We get a corresponding cubical diagram in the homotopy category as follows. The space $S^1$ and its six subspaces define an honestly commutative diagram which is almost all of a commutative cube, except that it's missing the initial vertex. Let the initial vertex be a point $\ast$, which maps isomorphically to all three objects $\{0\}, \{1/3\}, \{2/3\}$.
(Sorry, I'm not really up to TeXing up the commutative diagram on MO today, it would be much easier to grasp.)
This diagram in the homotopy category is homotopy commutative. In fact, all the spaces are connected and the sources of nontrivial morphisms are contractible, so the diagram has no choice but to commute. The six homotopies all occur in contractible mapping spaces, so there is no $\pi_1$-indeterminacy. The hexagon maps to the mapping space $Map(\ast,S^1) \cong S^1$, and if you use the most obvious choices of homotopies then the hexagon maps to $S^1$ by a homotopy equivalence.
In this instance you can choose a witness to each commutativity diagram. The failure to rectify homotopy commutativity occurs here because your witnesses aren't telling stories that are compatible.