I am reading about fibred categories. After reading the definition of "vertical" morphism, I can imagine why they are named like that. What about "cartesian" morphisms? What is cartesian about them? I don't know much mathematics besides some basic abstract algebra.


Let $\mathcal C$ be a category. A Cartesian square is a commutative square$\require{AMScd}$

\begin{CD} A @>a>> B\\@VVbV @VVcV\\ C @>d>> D \end{CD}

that is universal with respect to all commutative squares having the same bottom right hand corner. This means that if we have any commutative square of the form

\begin{CD} F @>f>> B\\@VVgV @VVcV\\ C @>d>> D \end{CD}

then there is a unique morphism $h\colon F\to A$ such that $b\circ h=g$ and $a\circ h=f$. As far as I know, the reason that these diagrams are called Cartesian is that one special example is given by the Cartesian product $A\times B$ of two sets:

\begin{CD} A\times B @>\text{pr}_1>> A\\@VV\text{pr}_2V @VVV\\ B @>>> \{*\} \end{CD}

What has this to do with Cartesian morphisms? If $\mathcal C$ is a category, then we can define the arrow category $\mathcal C^{\to}$ whose objects are morphisms $A\xrightarrow{a}B$ in $\mathcal C$ and where the morphisms from $A\xrightarrow{a}B$ to $C\xrightarrow{d}D$ are commutative squares:

\begin{CD} A @>a>> B\\@VVbV @VVcV\\ C @>d>> D \end{CD}

The arrow category is a prototypical example of a fibred category (as long as enough fibre products exist in the base category $\mathcal C$). There is a natural functor $p\colon\mathcal C^{\to}\to\mathcal C$ sending $A\xrightarrow{a}B$ to $B$ and the commutative diagram above to its right hand column. With respect to this functor, the morphisms in $\mathcal C^\to$ called Cartesian are precisely the Cartesian squares. The terminology has carried over to general fibred categories.

  • $\begingroup$ Nice answer. (I guess that, in the last paragraph, « sending $A\xrightarrow{a}B$ to $a$ » should be read « sending $A\xrightarrow{a}B$ to $B$ », probably). $\endgroup$ – Watson Mar 11 '18 at 17:11
  • $\begingroup$ @Watson Thanks! $\endgroup$ – John Gowers Mar 11 '18 at 17:25

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