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Sorry, I am new to category theory (actually all fields of math...). When I was learning the concept of monoid in college, the identity element is roughly defined as "an element e of set which satisfy $e * a = a = a * e$, for every a in that set". But when I am learning this concept again for understanding monad. The identity element of monoid is defined as a morphism (see here). Could anyone please explain why these two definitions of monoid (in particular the identity of monoid) are equivalent. Thanks!

EDIT:

Some other definition that represents the identity of monoid as morphism:

A definition from a math website

Another definition in stackoverflow (in the highest vote answer)

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  • $\begingroup$ Can you please post the exact definition of monoid and identity element you're referring to? I can probably answer this question given more details, but I can't extrapolate what definition you're trying to clarify from what you've written. $\endgroup$ – user231101 Sep 23 '16 at 16:15
  • $\begingroup$ The identity element "e" is such $e*a=a=a*e$. Compare your definition of the identity element. $\endgroup$ – amWhy Sep 23 '16 at 16:16
  • $\begingroup$ Hi @MikeHaskel, sorry I am not familiar with latex notation(it's my first time to post in this site). I have attach a link to wikipedia page where the identity of monoid is defined as a natural transformation. Thank you for you help. $\endgroup$ – Lifu Huang Sep 23 '16 at 16:27
  • $\begingroup$ In the linked definition, the unit is a morphism, not a natural transformation. Did you misread it, did you have a different definition in mind, or do you not see the difference between the two? $\endgroup$ – user231101 Sep 23 '16 at 16:43
  • $\begingroup$ @MikeHaskel the definition in the link is what I want to know. I have edited my question. I just realize that the reason why I thought identity as a natural transformation is that I read a definition somewhere else talking about monad. So could you please explain why this definition (identity as morphism) is equivalent to the one I learned in college? Thanks! $\endgroup$ – Lifu Huang Sep 23 '16 at 16:57
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The definition you linked to is doing more than just defining a monoid. It's actually defining a monoid object in a (monoidal) category. The usual notion of a monoid is a monoid object in $\operatorname{Set}$. Let's see how the definition plays out in that context.

According the the linked definition, the identity is a morphism from $1$—the unit object of the monoidal category—into the object of the monoid, $M$. In $\operatorname{Set}$, the unit object is any one-element set. The only data associated with a function from a one-element set to $M$ is knowledge of where the function sends that one element. That is, a function from a one-element set to $M$ is essentially just an element of $M$, so the definition aligns with the usual one.

The reason the general definition is phrased in terms of morphisms is because, in a general monoidal category, there might not be a notion of an "element." A morphism from the unit object to $M$ will still make sense, however.

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