A question on my Category Theory examples sheet is essentially as follows:

We define a degenerate monad $\mathbb{T} = (T, \eta, \mu)$ on $\mathbf{Set}$ to be one which has a non-monic unit; equivalently, there is no $\mathbb{T}$-algebra with more than one element. How many degenerate monads are there on $\mathbf{Set}$?

I've shown that every pair $( \{ a \}, \alpha: T \{ a \} \to \{ a \})$ is already an algebra, whatever the monad is (because $\{ a \}$ is terminal, so the "I am an algebra" commuting diagrams automatically commute).

Therefore, there are only two possible Eilenberg-Moore categories we could have made: "every one-element set", and "every one-element set, and also the empty set". That latter occurs iff $T \emptyset = \emptyset$.

There is the theorem which states that "for any $\mathbf{Set}$-monad $\mathbb{T}$, the forgetful functor $\mathbf{Set}^{\mathbb{T}} \to \mathbf{Set}$ has a left adjoint, and the adjunction induces $\mathbb{T}$"; from this I deduce that if we know the category of algebras, we have already determined $\mathbb{T}$ up to isomorphism.

Therefore, since we have only at most two options for the Eilenberg-Moore category, we must also only have at most two options for $\mathbb{T}$.

How can we determine whether these two options are indeed Eilenberg-Moore categories (and therefore whether we can induce monads from them)? All my theorems are about showing things about E-M categories, not about showing that something is an E-M category; the only constructive thing I can see is that the E-M category is initial over the category of adjunctions inducing $\mathbb{T}$, and I don't see how to apply that at all (and suspect it's not applicable).

I think that the one-point sets, together with the empty set, form the category of algebras over the monad $(T, \eta, \mu)$ where $T: A \to \{ a \}$, $\eta: \{ a \} \ni a \mapsto a \in A$, $\mu: \{ a \} \to \{ a \}$, and we've used super-industrial Choice to pick an element $a$ of every set $A$, and $T(\emptyset) = \emptyset$. However, this is something of a rabbit out of a hat.

  • $\begingroup$ Look up the monadicity / Barr-Beck theorem. $\endgroup$ – Qiaochu Yuan Nov 25 '15 at 22:19
  • 1
    $\begingroup$ Just compute directly. You know what the monad must be, so why not show that you have a monad? $\endgroup$ – Zhen Lin Nov 25 '15 at 22:34

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