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.

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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.