For each $R$ an object of a cartesian closed category, there is a monad $\mathrm{Cont}_R(A) = [[A,R],R]$, the continuation monad.

If $M$ is a strong monad, we can find for each pair of objects $A$ and $B$ a morphism $MA \times [A,MB] \to MB$, by using strength and then evaluating the exponential and then joining the result, so we can find a morphism $MA \to [[A,MB],MB]$, i.e. $MA \to \mathrm{Cont}_{MB}(A)$.

I think this is a monad morphism: I verified that the two ways of getting from $A$ to $[[A,MB],MB]$ were equal, drew the diagram for the commutation with $\mu$ and got a headache. In any case I believe that it works out fine.

So, roughly speaking we have a monad morphism from any (strong) monad to the continuation monad, which prompts me to say "aha! we have a terminal object!". However, there are two problems:

  • The monad morphism is $M \Rightarrow \mathrm{Cont}_{MB}$; crucially, the codomain depends on $M$, so there's not one single terminal object candidate to point to.
  • I wouldn't know where to even start proving that (or if!) this morphism is unique.

So, what's going on here, categorically speaking? Is this phenomenon actually expressible as a suitable universal property?

  • 1
    $\begingroup$ The terminal monad is actually, funnily enough, the constant $1$. $\endgroup$
    – Zhen Lin
    Jul 6, 2014 at 18:21
  • $\begingroup$ That's not so surprising, I suppose. But there's still something going on here. $\endgroup$ Jul 6, 2014 at 22:50
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    $\begingroup$ I have a hunch that given a monad $T$ and an object $R$, a monad morphism $T \rightarrow \mathrm{Cont}_R$ is the same thing as a $T$-algebra structure on the object $R$. Thoughts/comments? $\endgroup$ Dec 17, 2016 at 23:16
  • $\begingroup$ I think $\operatorname{Cont}_R$ along with $eval : \operatorname{Cont}_R(R) \to R$, $f \mapsto f (\operatorname{id}_R)$, would be terminal in the category of monads $M$ with a distinguished $M$-algebra structure on $R$. Similar to the observation from goblin's comment. $\endgroup$ Feb 22, 2018 at 20:02

1 Answer 1


One observation I have made since asking this question: $\mathrm{Cont}_{MB}$ frequently has non-trivial automorphisms, and so cannot be a genuine terminal object. Any automorphism of $M$ itself, say $\alpha : M \Rightarrow M$, gives rise to an automorphism which I'll call $\overline{\alpha}$ of $\mathrm{Cont}_{MB}$ via \begin{aligned} \overline{\alpha} \ &: \mathrm{Cont}_{MB} \Rightarrow \mathrm{Cont}_{MB} \\ \overline{\alpha}_C \ &: [[C, MB], MB] \to [[C, MB], MB] \\ \overline{\alpha}_C(f) \ &= k \mapsto \alpha_B^{-1}(f(\alpha_B \circ k)) \end{aligned} I admit I haven't gone through the details, but it seems intuitively like it must work.


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