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Specifically, are there "nice" (ie, not too obscure or contrived) examples of families of objects, where say for each objects $A,B,C$ in the family, the canonical isomorphism from $A\rightarrow C$ is not the composition of the canonical isomorphisms from $A\rightarrow B$ and $B\rightarrow C$?

To illustrate, here's a non-example: The fundamental groups of a space with different base-points are all canonically isomorphic up to inner automorphisms, so given a space $X$, the fundamental groups $\pi_1(X,x)$ as $x$ varies over $X$, together with the conjugacy classes of these 'canonical isomorphisms', form a category.

Will Chen
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  • Conjugacy classes of isomorphisms do not form a category in any really sensible way. – Todd Trimble Oct 23 '14 at 03:10
  • I claim that if this ever happens to you then at least one of the isomorphisms in question shouldn't have been considered canonical. @Todd: really? Isn't the OP just working in the homotopy category of groups (equivalently, the homotopy category of Eilenberg-MacLane spaces)? – Qiaochu Yuan Oct 23 '14 at 04:37
  • @QiaochuYuan I'm not seeing it. For a simple example of what I meant, take a groupoid with one object, say with automorphism group $S_4$. There are five conjugacy classes; what is the group structure on the set of conjugacy classes? – Todd Trimble Oct 23 '14 at 05:12
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    @Todd: that isn't what I thought either you or the OP were referring to. Here is the category I am referring to: the objects are groups, and the morphisms $G \to H$ are conjugacy classes of homomorphisms $f : G \to H$, where two homomorphisms $f_1, f_2$ are conjugate if there exists some $h \in H$ such that $h f_1(g) h^{-1} = f_2(g)$ (the same as saying that they're naturally isomorphic as functors). This is a perfectly well-defined category which deserves to be called the homotopy category of groups; in particular it is equivalent to the homotopy category of Eilenberg-MacLane spaces. – Qiaochu Yuan Oct 23 '14 at 05:20
  • Moreover, if $X$ is a path-connected space, the fundamental groups $\pi_1(X, x)$ naturally give a diagram in this category: there is an isomorphism $\pi_1(X, x_0) \to \pi_1(X, x_1)$ for every choice of path from $x_0$ to $x_1$, and choosing a different path by composing by some element of $\pi_1(X, x_1)$ changes this isomorphism by a conjugation, so indeed the resulting isomorphism is unique in the homotopy category. Said another way, $\pi_1(X)$ (without a choice of basepoint) gives a well-defined functor from path-connected (but not pointed) spaces to the homotopy category of groups. – Qiaochu Yuan Oct 23 '14 at 05:23
  • @QiaochuYuan That makes it much clearer; thanks. – Todd Trimble Oct 23 '14 at 05:32
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    Would you consider parallel transport along a prescribed curve in a Riemannian manifold with nontrivial holonomy to be an example? – Paul Siegel Oct 23 '14 at 05:39

4 Answers4

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Maybe this is kind of contrived, but any single object with a canonical automorphism that is not the identity is an example of this. In fact, any other example must in some sense include an example like this, if you consider the composition of $A \to B \to C$ with the inverse of $A \to C$ a "canonical" automorphism of $A$.

Eric Wofsey
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Here is an example. Consider the family of rotated copies of the unit square in the plane $\mathbb{R}\times\mathbb{R}$, rotated by some angle about the origin. If one such copy is rotated only a little from a second, then it seems the canonical isometry to bring them into alignment would be to rotate the first through the smallest possible angle to bring it into alignment with the second. But of course, the composition of many such small angle rotations (or even just two of sufficient small size) would add up to a large angle, and so the composition of these canonical isometries is no longer canonical.

(I suppose that one should say here that if the figures are rotated by exactly $45^\circ$, then either of the two minimal rotations should count as canonical.)

Joel David Hamkins
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    One should say here that if the figures are rotated by exactly $45^{\circ}$, then neither of the two minimal rotations should count as canonical! This is essentially the issue that comes up in writing down, say, branches of the logarithm; you need to pick a "branch cut." – Qiaochu Yuan Oct 23 '14 at 04:33
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    In the first paragraph I thought that the unit square would be $[0,1]^2$ but then I realized that you had $[-1,1]^2$ in mind. – Dirk Oct 23 '14 at 08:42
  • @Dirk Yes, I meant to rotate the square about its center. – Joel David Hamkins Oct 23 '14 at 14:35
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Let $X=\mathbb{A}_{\mathbb{C}}^1\backslash\{0\}$. The first algebraic de Rham cohomology of $X$ is 1-dimensional, and has a canonical generator $\frac{dz}{z}$ (or any other 1-form with residue 1). The first Betti cohomology is also 1-dimensional, and has a canonical generator taking a 1-cycle to its winding number around 0. Over $\mathbb{C}$ there is a canonical isomorphism $H^1_{dR}(X)\to H^1_B(X,\mathbb{C})$, but it takes the generator to $2\pi i$ times the generator.

If we wanted to say this in terms of isomorphisms, we could consider the three vector spaces $\mathbb{C},H^1_{dR}(X)$, and $H^1(X,\mathbb{C})$, and use the fact that an isomorphism from $\mathbb{C}$ to a 1-dimensional vector space is the same thing as the choice of a generator of that vector space.

Julian Rosen
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You don't really need to construct sophisticated mathematical examples to answer this. The problem is that the word canonical has at best a sociological meaning, not a mathematical one.

Are $X\times Y$ and $Y\times X$ "canonically isomorphic" by the switching map?

In some contexts it may be reasonable to say so, in order to avoid bureaucratic notation. However, once $X$ is non-trivial and $Y=X$ you have a non-trivial group.

All that word canonical means is "I'm too lazy to write down the definition".

Paul Taylor
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  • Canonical really means "I don't have to bother writing down the definition because there's only one possibility which doesn't involve making arbitrary choices". Thus $X \times Y$ is canonically isomorphic to $Y \times X$, but a finite dimensional vector space is not canonically isomorphic to its dual. – Paul Siegel Oct 23 '14 at 17:37