I'm coming across the following 'commuting' diagram a lot in my work, and I think it should have a neat categorical formulation. But I can't work it out for myself, and don't know what too google for. Maybe somebody can give me a pointer?

Assume I have 4 functions/morphisms.

  • $f : A \rightarrow B$.
  • $down : A \rightarrow A'$.
  • $up : B' \rightarrow B$.
  • $g : A' \rightarrow A' \oplus B'$.

Here $\oplus$ is the disjoint sum of sets. I want to say the following diagram 'commutes':

Should commute but doesn't 'typecheck'

but it doesn't 'typecheck' because the codomain of $g$ doesn't match the domain of $up$. However, it does 'commute eventually' in the following sense. For all $a \in A$ the following holds. There exist $a_0, ..., a_{n-1} \in A'$ and $b' \in B'$ such that:

  • $down(a) = a_0$,
  • $g(a_0) = inl(a_1)$,
  • $g(a_1) = inl(a_2)$,
  • ...
  • $g(a_{n-2}) = inl(a_{n-1})$,
  • $g(a_{n-1}) = inr(b')$, and
  • $up(b') = f(a)$.

Here $inl(a)$ means $a$ is in the 'left' part of the sum $A' \oplus B'$ while $inr(b)$ means $b$ is in the 'right' part of $A' \oplus B'$. In other words, for the diagram to 'commute', $g$ should be applied until it yields something of type $B$. In my application, this will always eventually happen. If we were sloppy and ignored the $inl/inr$ we could write: $\forall a\in A. \exists n. f(a) = up ( g^n(down(a)))$.

What's a good categorical rendering of this? Some abstract conception of iteration or recursion? In which parts of mathematics do similar constructions crop up?

  • 1
    $\begingroup$ So $g(\text{down}(A))$ is contained in $B'$ ? $\endgroup$ Jun 16, 2015 at 12:32
  • 1
    $\begingroup$ @StefanHamcke Not necessarily. The 'transitive closure' of $g^*$ applied to $down(A)$ is in $B'$. For each individual $a \in A'$, the number of times I need to apply $g$ before I arrive in $B'$ is typically different. $\endgroup$ Jun 16, 2015 at 12:38
  • 1
    $\begingroup$ I see. Is this true for all point in $A'$, or just for those in the image of $A$?. I mean, for each $a\in A'$, can you apply $g$ some finite amount of times until you arrive in $B'$? $\endgroup$ Jun 16, 2015 at 12:46
  • 1
    $\begingroup$ @StefanHamcke In my applications this is true for all of $A'$, It's very important that this is true. $\endgroup$ Jun 16, 2015 at 12:47

1 Answer 1


Since you're interested in considering objects that have elements I'll assume we are working with $\mathbf{Sets}$ the category of sets and functions.

From the function $g \colon A' \to A' \oplus B'$ and from $inr \colon B' \to A' \oplus B'$ you can obtain the function $\bar g \colon A' \oplus B' \to A' \oplus B'$. By definition this function is such that

  • for every $a \in A'$ we have $\bar g(inl(a))=g(a)$;
  • for every $b \in B'$ we have $\bar g(inr(b))=b$.

Now using your notation we have that $inl(a_1)=\bar g(inl(a_0))$ and an easy induction shows that for every $n=1,\dots,n$ we have that $inl(a_n)=\bar g^n(inl(a_0))$. So your requirement that the sequence of the $a_n$'s eventually falls in $B$ can be expressed as

$$\forall a \in A'\ \exists b \in B'\ \exists n \in \mathbb N\ \bar g^n(inl(a))=inr(b)\ .$$

This should imply that $B'$ is the colimit of the diagram

$$A' \oplus B' \stackrel{\bar g}{\longrightarrow}A' \oplus B' \stackrel{\bar g}{\longrightarrow}A' \oplus B' \stackrel{\bar g}{\longrightarrow}A' \oplus B' \stackrel{\bar g}{\longrightarrow}A' \oplus B' \stackrel{\bar g}{\longrightarrow}A' \oplus B' \dots$$ where the colimit maps $(in^g_i \colon A' \oplus B' \to B')_{i \in \mathbb N}$ should send every $inl(a)$ into the eventual limit of the sequence $\{\bar g^n(inl(a))\}_{n \in \mathbb N}$ and every $inr(b)$ into $b$ (itself).

Now you could express your condition via the commutativity of the following diagram $$ \require{AMScd} \begin{CD} A @>{f}>> B \\ @V{down}VV @A{up}AA\\ A' @>{in^g_1 \circ inl}>> B' \end{CD} $$ where $inl$ is the embedding of $A'$ into $A' \oplus B'$ while $in^g_1$ is the first structure map of the above mentioned colimit diagram $(A' \oplus B' \stackrel{in^g_i}{\longrightarrow} B')_{i \in \mathbb N}$.

Hope this helps.


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