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

How can the two diagrams in (2) be written as identity equations?



In any diagram: pick two objects $x,y$, list all 'morphism paths' that start in $x$ and end in $y$. All these compositions ought to be equal. This is rather direct for commuting squares and triangles, since we have few paths to consider. Here, we have

$$ \mu \circ T\mu = \mu \circ \mu T $$


$$\mu \circ \eta T = \mu \circ T\eta = 1_T.$$

  • $\begingroup$ Thanks. Do the identity equations not depend on the operands: $T^3, T^2, T, IT, TI$? $\endgroup$ – Tim Jul 24 at 2:04
  • $\begingroup$ I dont' know what do you mean by operand here. These are the domains and codomains of each arrow, they implicit in their composition. More importantly, $T$ is fixed: the mondad is the $3$-uple $(T,\mu,\eta)$. $\endgroup$ – Guido A. Jul 24 at 2:09
  • $\begingroup$ Why do your identity equations not specify what $\mu, T\mu, \mu T, \eta T, T \eta$ apply to $\endgroup$ – Tim Jul 24 at 2:16
  • $\begingroup$ That's implicit from the definition: the multiplication is an arrow $\mu : T^2 \Rightarrow T$ and the unit is an arrow $\eta : 1_T \Rightarrow T$. From these fact one can deduce the (co)domain of all the other arrows. $\endgroup$ – Guido A. Jul 24 at 2:44
  • $\begingroup$ Thanks. Why is $\mu$ called multiplication and $\eta$ unit? math.stackexchange.com/questions/3302168/… $\endgroup$ – Tim Jul 24 at 11:23

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