Categories for the Working Mathematician says

Definition. A monad $T= \langle T, \eta, \mu\rangle $ in a category $X$ consists of a functor $T: X \to X$ and two natural transformations

$$\eta : I_X \Rightarrow T, \mu : T^2 \Rightarrow T $$

which make the following diagrams commute

enter image description here

Given $\eta : I_X \Rightarrow T$, i.e. $\eta$ is a natural transformation from identity functor $I_X$ to functor $T$, I sometimes have the illusion that $\eta$ should just be $T$, because $\eta(I_X)=T$ and $T \circ I_X = T$. So is $\eta$ the same as $T \circ$?

I wonder how to explain whether that is correct or incorrect?


  • 6
    $\begingroup$ The question as it is does not make much sense: $T$ is a functor and $\eta$ is a natural transformation, you can equate potatoes and carrots... Go through some example of monads to get some insights on the unit $\eta$. For example, in the case of the free monoid monad $T: \mathbf{Set} \to \mathbf{Set}$, the unit $\eta_X : X \to TX$ at component $X$ is the map that sends an element $x$ to the word $x$ of length $1$. $\endgroup$ – Pece Jul 17 at 11:31
  • $\begingroup$ Thanks. I couldn't find free monoid monad in my books. Which books do you know that describe it? $\endgroup$ – Tim Jul 20 at 17:13
  • $\begingroup$ @Pece I have read more about the free monad, and also have just added more about my confusion. Hope you could enlighten more $\endgroup$ – Tim Jul 24 at 11:36
  • 2
    $\begingroup$ The expression $\eta(I_X)$ doesn't make much sense. For any object $c \in X$ you get a map $\eta_c: c \rightarrow Tc$, but it is not at all clear what is the meaning of $\eta(I_X)$. However, as Pece told you, it doesn't make sense to equate a functor and a natural transformation, since these are fundamentally different things. $\endgroup$ – Charlie Jul 24 at 11:57
  • $\begingroup$ @Charlie Isn't $\eta: I_X\to T$ a natural transformation transforming $I_X$ to $T$? So doesn't $η(I_X)$ make sense? $\endgroup$ – Tim Jul 24 at 12:03

The equation $\eta(I_X) = T$ doesn't really make sense. A natural transformation isn't a machine that takes in functors and produces new functors, its a machine that takes in objects $A \in X$ and produces morphisms $\eta_A: A \to T(A)$.

For a general functor $T$, $A$ and $T(A)$ may have no relationship whatsoever, so in that sense the morphisms that make up $\eta$ are a non-trivial piece of data - $\eta$ is not "just" equal to any one thing in particular.

A simple example to play with is a category $X$ coming from a monoid $M$, considered as a category with one object $*$ such that $Hom(*,*) = M$. Then functors $T$ are simply monoid endomorphisms $T \in End(M)$. A natural transformation $\eta: I_X \to T$ is then just a singleton collection $\eta = \{\eta_*\}$ of one morphism for the object $*$, which is by definition some element $n \in M$. The naturality condition says that for all $x \in M$, one has $T(x)\cdot n = n\cdot x$.

In the special case that $T = I$ is the identity functor, the equation becomes $xn = nx$ so that choices of $\eta$ correspond to central elements $n \in Z(M)$. To extend this example to an actual monad you need to assume $n$ has a left inverse which then plays the role of $\mu$ - this is because the righmost diagram requires $x = (\mu_* \circ \eta_*)(x)$ for all $x \in M$.)

NOTE: Sorry I realize now that $\eta_*, \mu_*$ looks kind of like "pushforward" notation, but here I mean the natural transformation evaluated at the object $*$ of a one-object category.


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