A standard example for demonstrating the need for genuinely weak ncategories is that a weak 3category with unique 0 and 1cells amounts to the same thing as a braided monoidal category (by an EckmannHilton argument), but were one to use a strict 3category instead, this would automatically become a fortiori symmetric. In trying to get a better intuition for "the right notion" of weak categories (this is still unsettled, right?), I was wondering if anyone could give me a good intuition for what step in the argument for symmetry in the strict case we want to fail in the weak case and why. [I suppose this amounts to more generally giving a good intuition for what higherdimensional coherence isomorphisms we should refrain from demanding to exist in the definition of weak categories, and why, but this example seems in particular like a usefully illustrative introductory context]
Here is a nice pictorial proof of the EckmannHilton argument in a higher category (made by Eugenia Cheng). One way to read this is as a proof that in a weak 2category with one 0cell and one 1cell, the composition of 2cells is commutative: in this case each diagram is equal to the two diagrams next to it, either by a unit law or by an interchange law.
Alternately, we can read it as a proof that in a weak 3category with unique 0 and 1cells, the composition on 2cells is braided. In this case each diagram is isomorphic to those next to it, via either a unit isomorphism or an interchange isomorphism. The braiding is the composite isomorphism along the top half (say), and the fact that it isn't a symmetry arises because the composite all the way around need not be the identity.
Of course, if both interchange and unit isomorphisms were to be identities, the braiding would be a symmetry. It is known in dimension 3 that it's actually good enough (i.e. things don't collapse and become too strict) if either the unit or interchange law is weak, and the other is strict. If interchange is weak but the unit is strict, we get Graycategories as in the original GordonPowerStreet paper on tricategories, while if the interchange is weak but the unit is strict, this is the subject of Simpson's conjecture as mentioned by David (for the n=3 case, see here). Note that associativity never enters the picture at all, and at least in dimension 3 it can also be made strict.
There isn't really a "one step" that we want to fail. In a "fully weak" higher category, all of the associativity, unit, and interchange laws would hold only weakly. It so happens that such fully weak categories can be "semistrictified" to make some, but not (in general) all, of these laws hold strictly, while still being equivalent to the original category in a suitable sense. Such semistrictification can often be technically useful, but I don't regard it as really of foundational importance.
Regarding "intuition for what higherdimensional coherence isomorphisms we should refrain from demanding to exist," I think it's misleading to view this example in that way. There isn't really a coherence isomorphism that we're refraining from demanding to existin a weak 3category, it's still the case that all coherence isomorphisms exist and all "formally describable" diagrams commute. Rather, what's happening is that accidentally, if we happen to be considering cells whose source and target are both identities, then there are some structural isomorphisms that we happen to be able to compose in a way "unforeseen" by the general theory of weak ncategories. Therefore, in this particular case, the assertion of that general theory that "all diagrams commute" doesn't apply, since the diagram we're looking at is only welldefined by accident. This is kind of vague, but it can be made precise by the theory of contractible globular operads, where it is true that "all diagrams commute" in the formal, operadic, sense, but in some particular algebra over such an operad, there may be accidental "composites" which do not commute. See also this question.

$\begingroup$ I'm not sure how to read the diagram you linked to. It looks like a diagram of a plainvanilla EckmannHilton argument, except for the $l$s and $l^{1}$ pieces, and I'm not sure what those are. That having been said, I <i>think</i> I understand anyway. At least, the last paragraph above seems like what I was thinking the answer was anyway (as for why there's no demand for a coherence isomorphism between the path all the way around and the identity). But let me make sure I understand: (Question coming in a followup comment with more characters left...) $\endgroup$ – Sridhar Ramesh Feb 11 '10 at 6:56

$\begingroup$ Is the idea here analogous to the fact that, in some sense, the theory of ordinary categories (which I want to describe using a multicategory of some sort, but can't quite, but suppose I could if I had the multicategory live in some fancy category other than Set) sees one and only one function from Hom(A, B) x Hom(B, C) to Hom(A, C) for objects A, B, and C [binary composition], even though, for any particular actual objects in an actual category, there may be many nonequivalent functions of this type which can be built out of the structure of a category $\endgroup$ – Sridhar Ramesh Feb 11 '10 at 7:12

$\begingroup$ (e.g., if A, B, and C happened to all be identical, in addition to composing the left and right morphisms, there would also be simply projection the left morphism, projecting the right morphism, producing the square of the left morphism, etc.)? That is, the theory of categories doesn't concern itself with such cases as where the three objects in binary composition all happen to accidentally line up; it can't see all these functions from Hom(A, A) x Hom(A, A) to Hom(A, A) for objects A, and so it doesn't impose coherence equalities between them. $\endgroup$ – Sridhar Ramesh Feb 11 '10 at 7:12

$\begingroup$ Is it something like a higherdimensional analogue of that same phenomenon? (Man, these character limits are brutal... this and the last two posts were meant to be one single comment) $\endgroup$ – Sridhar Ramesh Feb 11 '10 at 7:13

1$\begingroup$ Sridhar, the diagrams in the EckmannHilton clock live in a bicategory, and the 2cells marked $\ell$ are unit coherence isomorphisms. The setup is this: we've got some object A of some bicategory, and we've got 2cells alpha and beta from 1_A to 1_A. The 2cell $\ell$ is the coherence isomorphism $1_A \circ 1_A \to 1_A$. $\endgroup$ – Tom Leinster Feb 11 '10 at 13:03
One thought along these lines is Simpson's conjecture (proved for $n$categories up to $n=3$). Essentially this states that a weak $n$category is equivalent to one where everything is strict except for the identities. Since one of the assumptions in EckmannHilton is that there is a unit for both multiplications, we lose the result that the product is abelian/symmetric.
Another approach is along the lines of weakening interchange (another assumption of EckmanHilton, namely that the products commute with each other), which is what happens in Graycategories  categories enriched over 2Cat with the Gray tensor product (as opposed to the cartesian product). Graycategories are known to model all weak 3categories. Crans has taken this further and constructed an analogous tensor product on GrayCat, but I'm not certain if enriching using this tensor product gives weak 4categories.

1$\begingroup$ A reference for the first part of this is to work of Joachim Koch with André Joyal. $\endgroup$ – Tim Porter Feb 11 '10 at 8:27

$\begingroup$ Of course  was being a bit sloppy in that regard. $\endgroup$ – David Roberts Feb 12 '10 at 4:04