I am struggling to understand the definition of a final functor given on page 217 of MacLane's "Categories for the Working Mathematician." It says that a functor $L \colon J' \rightarrow J$ is final if for each $k \in J$ the comma category $(k/L)$ is nonempty and connected. Let's suppose that $L$ is an inclusion, and $J$ is the pushout data $$ B \leftarrow A \rightarrow C $$ If $L$ is simply the identity, it does not seem to satisfy the conditions of being final. That is, the comma category under the $A$ is not connected as far as I can tell. But surely the identity functor is final?

I'm sure I am misunderstanding something and would appreciate if someone could set me straight.

  • $\begingroup$ Why wouldn't it be connected? There's a morphism connecting $id_A:A\to A$ to each of the other two objects in $(A\downarrow L)$. $\endgroup$ Mar 20, 2020 at 17:30
  • 1
    $\begingroup$ Wow, do I feel silly...yes, I see it now. I had it in my head that you'd need a morphism from $B$ to $C$ or the other way around. Total brainfart. Thank you for clarifying! $\endgroup$
    – user761471
    Mar 20, 2020 at 17:50
  • $\begingroup$ A follow-up question as well. Why would the inclusion of the category with just the objects $B \ A \ C$ (removing the two original morphisms) not be final? $\endgroup$
    – user761471
    Mar 20, 2020 at 18:18
  • $\begingroup$ If you take $J'$ to be the category with the same objects but only the identity morphisms, then $A/L$ also has only identity morphisms, but it has three objects $A\to A$, $A\to B$, $A\to C$ so it is not connected. $\endgroup$
    – Arnaud D.
    Mar 20, 2020 at 18:28
  • $\begingroup$ Thank you (both), that has cleared everything up! $\endgroup$
    – user761471
    Mar 20, 2020 at 19:57

1 Answer 1


One way i build intuition is from when considering colimits indexed by directed posets.

If $P$ is some directed poset and $P'\subseteq P$ is some subset. Then the inclusion $i:P'\to P$ is a final functor if for all $p \in P$ there exists some $p'\in P'$ such that $p\leq p'$. Hence it is sufficient to compute colimits indexed by $P'$.

This plays nice with my intuition since in nice situations such colimits should be fought of as unions in which thise should be very clear.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy