Canonically isomorphic but not equal In mathematics, we have many objects that are canonically isomorphic but not equal on the nose.
For example let $V$ and $W$ be vector spaces. Then $V\otimes W$ and $W\otimes V$ are canonically isomorphic but not equal. But normally we do not care about it and deal them as equal.
I am wondering if there is a situation that this subtlety is very important. 
Thanks.
 A: Sometimes, two objects are equal but canonically isomorphic via a non-trivial isomorphism. For instance, one might be tempted to replace a category with its skeleton, so that every isomorphism is an automorphism – but this does not necessarily force "canonical" isomorphisms to be the identity! See the closing remarks in [Categories for the working mathematician, Ch. VII, § 1]:

One might be tempted to avoid all this fuss with $\alpha$, $\lambda$, and $\rho$ by simply identifying all isomorphic objects in $B$. This will not do, by the following argument due to Isbell. Let $\textbf{Set}_0$ be the skeleton of the category of sets; it has a product $X \times Y$ with projections $p_1$ and $p_2$ as usual. If $D$ is a (the) denumerable set, then $D = D \times D$, and both projections of this product are epis $p_1, p_2 : D \to D$. Now suppose that the isomorphism $\alpha : X \times (Y \times Z) \to (X \times Y) \times Z$, defined as usual to commute with the three projections, were always the identity; it is then the identity for $X = Y = Z = D$; since $\alpha$ is natural, $f \times (g \times h) = (f \times g) \times h$ for any three $f, g, h : D \to D$. But $\times$ on functions is defined in terms of the projections $p_1$ and $p_2$ above, so
  $$f p_1 = p_1 (f \times (g \times h)) = p_1 ((f \times g) \times h) = (f \times g) p_1$$
  and $p_1$ is epi, so $f = f \times g$. The corresponding argument with $p_2$ gives $f \times g = g$, hence $f = g$ for any $f, g : d \to D$, an absurdity. A similar argument applies to the skeleton of $\langle \textbf{Ab}, \otimes, \cdots \rangle$.

