# Given $\eta : I_X \Rightarrow T$, is $\eta$ just $T \circ$?

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 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?

Thanks.

• 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 $1$. – Pece Jul 17 at 11:31
• Thanks. I couldn't find free monoid monad in my books. Which books do you know that describe it? – Tim Jul 20 at 17:13
• @Pece I have read more about the free monad, and also have just added more about my confusion. Hope you could enlighten more – Tim Jul 24 at 11:36
• 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. – Charlie Jul 24 at 11:57
• @Charlie Isn't $\eta: I_X\to T$ a natural transformation transforming $I_X$ to $T$? So doesn't $η(I_X)$ make sense? – 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.