MHB Monoids as Categories .... Awodey Section 1.4, Example 13 ....

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Example Section
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.4 Examples of Categories ...

I need some help in order to fully understand some aspects of Section 1.4 Example 13 ...

Section 1.4 Example 13 reads as follows:https://www.physicsforums.com/attachments/8343
View attachment 8344In the above text by Awodey we read the following:

" ... ... But also for any set $$X$$ the set of functions from $$X$$ to $$X$$ , written as

$$\text{HOM}_\text{Sets} (X, X)$$

is a monoid under the operation of composition. More generally, for any object $$C$$ in any Category $$C$$, the set of arrows from $$C$$ to $$C$$, written as

$$\text{HOM}_C (C, C)$$

is a monoid under the composition operation of $$C$$. ... ... "
I am slightly unsure regarding how to interpret the objects and arrows of $$\text{HOM}_\text{Sets} (X, X)$$ and $$\text{HOM}_C (C, C)$$ when viewed as categories ... ... ?My interpretation of $$\text{HOM}_\text{Sets} (X, X)$$ is that the single object is $$X$$ and the arrows are the functions for $$X$$ to $$X$$ ... ... is that correct?My interpretation of $$\text{HOM}_C (C, C)$$ is that that the single object is $$C$$ and the arrows are the arrows from $$C$$ to $$C$$ ... ... is that correct?Help will be appreciated ...

Peter
 
Last edited:
Physics news on Phys.org
Peter said:
I am slightly unsure regarding how to interpret the objects and arrows of $$\text{HOM}_\text{Sets} (X, X)$$ and $$\text{HOM}_C (C, C)$$ when viewed as categories ... ... ?
I think the author suggests viewing $$\text{HOM}_\text{Sets} (X, X)$$ as a monoid in the usual sense of an algebraic structure rather than as a category. The fact that it is an algebraic monoid is obvious since function from $X$ to $X$ are elements and composition is the operation. But yes, since every monoid is a category, $$\text{HOM}_\text{Sets} (X, X)$$ can be viewed as a category as well.

Peter said:
My interpretation of $$\text{HOM}_\text{Sets} (X, X)$$ is that the single object is $$X$$ and the arrows are the functions for $$X$$ to $$X$$ ... ... is that correct?
We can make anything to be the single object, for example, the symbol $*$: it does not matter. The arrows are indeed functions from $X$ to $X$, viewed as arrows from and to that single object.

Peter said:
My interpretation of $$\text{HOM}_C (C, C)$$ is that that the single object is $$C$$ and the arrows are the arrows from $$C$$ to $$C$$ ... ... is that correct?
It can be viewed this way.
 
Evgeny.Makarov said:
I think the author suggests viewing $$\text{HOM}_\text{Sets} (X, X)$$ as a monoid in the usual sense of an algebraic structure rather than as a category. The fact that it is an algebraic monoid is obvious since function from $X$ to $X$ are elements and composition is the operation. But yes, since every monoid is a category, $$\text{HOM}_\text{Sets} (X, X)$$ can be viewed as a category as well.

We can make anything to be the single object, for example, the symbol $*$: it does not matter. The arrows are indeed functions from $X$ to $X$, viewed as arrows from and to that single object.

It can be viewed this way.
Thanks Evgeny ... Appreciate your help...

Peter
 
Evgeny.Makarov is right. $Hom(X,X)$ is not viewed as a category, here.

See Example 1.5.1 of Simmons for a better explanation of how to view a monoid as a category.
 
steenis said:
Evgeny.Makarov is right. $Hom(X,X)$ is not viewed as a category, here.

See Example 1.5.1 of Simmons for a better explanation of how to view a monoid as a category.
Thanks Steenis ... will check Simmons ...

Peter
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top