Consider the category Cat as a concrete category over Set $\times$ Set via the functor
U : Cat $\rightarrow$ Set $\times$ Set, defined by

U$(\mathbf A \xrightarrow{F} \mathbf B) = ($Ob$(\mathbf A)\xrightarrow{F_O}$ Ob$(\mathbf B)$ , Mor$(\mathbf A)\xrightarrow{F_M}$ Mor$(\mathbf B))$,

where $F_O$ is the restriction of $F$ to objects and $F_M$ is its restrictions to morphisms.

Cat is the category of all small categories in the sense that their objects and morphisms form sets (not classes).

It is clear that this concrete category is transportable. But is this category uniquely transportable?

A concrete category $(\mathbf A, U)$ over $\mathbf X$ is said to be (uniquely) transportable provided that for every $\mathbf A-$object A and every $\mathbf X-$isomorphism $UA\xrightarrow{k}X$ there exists a (unique) $\mathbf A-$object B with $UB=X$ such that $A\xrightarrow{k}B$ is an $\mathbf A-$isomorphism.

  • 3
    $\begingroup$ For all I know, the collocation "concrete category over $\mathbf{Set}\times\mathbf{Set}$" in this OP is very unusual. The term 'concrete category' is very usual of course, but that just means 'category $\mathsf{C}$ such that there is a faithful functor to $\mathsf{Set}$. (Cf. e.g. Riehl 2014, Def. 1.6.17) Would you mind explaining where you get this usage from? $\endgroup$ Oct 7, 2017 at 11:53
  • $\begingroup$ Data point for those who would like to write an answer: in [J. Adámek,J. Rosický,E. M. Vitale: Algebraic Theories: A Categorical Introduction to General Algebra. Cambridge University Press 2011], Example 13.17.2 gives the category $\Sigma$-Alg of all '$\Sigma$-algebras' (perhaps not needless to say for some readers, these are not the '$\sigma$-algbras' of measure theory) as an example of a uniquely-transportable concrete category. $\endgroup$ Oct 7, 2017 at 12:00
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    $\begingroup$ I'd say the answer is obviously 'yes', since the source, target, identity, and composition data that lift $X = (O, M)$ to a category structure (rendering $k$ an isomorphism of categories) is uniquely determined by "conjugating" the category structure on $A$ by the data of $k$. If this is not enough of a hint, then I guess I could write out the answer. $\endgroup$
    – Todd Trimble
    Oct 7, 2017 at 14:02
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    $\begingroup$ @PeterHeinig, I took this usage from “Abstract and Concrete Categories” book. There you will find the notion of concrete category over an arbitrary category $\mathbf X$ if underlying functor is faithful. $\endgroup$
    – A. G
    Oct 7, 2017 at 14:48
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    $\begingroup$ While The Joy of Cats may have idiosyncrasies, I wouldn't consider this one of them. The usage here is immediately understandable and sensible. $\endgroup$
    – Todd Trimble
    Oct 7, 2017 at 15:19

1 Answer 1


Let $X = (O, M)$ be a pair of sets, and let $A$ be a category with set of objects $O_A$ and set of morphisms $M_A$. Suppose given a pair $k = (k_0, k_1)$ of isomorphisms $k_0: O_A \to O$, $k_1: M_A \to M$. Let $s_A, t_A: M_A \to O_A$ be the source and target functions for $A$; then the only way to define the source and target functions $s, t: M \to O$ on the lifted structure over $X$ in such a way that $k$ preserves source and target data is by setting

$$s = k_0 \circ s_A \circ k_1^{-1}, \qquad t = k_0 \circ t_A \circ k_1^{-1}$$

(that's what I meant by "conjugating" in my comment).

It's the same idea to define the lifted structure $i, c$ for identity and composition. If $i_A, c_A$ are the identity and composition on $A$ and $k$ is to preserve identity and composition data, then there's only one way to define them: put

$$i = k_1 \circ i_A \circ k_0^{-1}, \qquad c = [(f, g) \mapsto (k_1 \circ c_A)(k_1^{-1} f, k_1^{-1} g)]$$

where $f, g$ is any composable pair in the lifted source-target structure over $X$.

Of course unique transport of structure via conjugation is a simple idea which applies much more generally, as hinted in Peter's second comment.

  • $\begingroup$ +1 for the last paragraph. $\endgroup$
    – HeinrichD
    Oct 30, 2017 at 13:32

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