# Understanding Monoids as Categories: Awodey Section 1.4 Example 13

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In summary, Steve Awodey's book Category Theory (Second Edition) provides a detailed and rigorous explanation of the concept of a functor. Awodey lays down three conditions that must be met in order for a functor to be a valid Functor. His monoid-homomorphism $h:M \rightarrow N$ meets all three of these conditions.
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I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.4 Examples of Categories ...

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

Section 1.4 Example 13 reads as follows:View attachment 8347
View attachment 8348
In the above text by Awodey we read the following:

" ... ...In detail, a homomorphism from a monoid $$\displaystyle M$$ to a monoid $$\displaystyle N$$ is a function $$\displaystyle h : M \to N$$ such that for all $$\displaystyle m,n \in M$$,$$\displaystyle h( m \bullet_M n ) = h(m) \bullet_N h(N)$$

and

$$\displaystyle h( u_M ) = u_N$$

Observe that a monoid homomorphism from $$\displaystyle M$$ to $$\displaystyle N$$ is the same thing as a functor from $$\displaystyle M$$ regarded as a category to $$\displaystyle N$$ regarded as a category. ... ... "I cannot see how (exactly and rigorously) the monoid homomorphism specified above fits the 3 conditions (a), (b) and (c) that Awodey lays down for a functor (see text below) ... ...

Can someone please demonstrate explicitly and rigorously how (exactly) the monoid homomorphism specified above fits the 3 conditions (a), (b) and (c) that Awodey lays down for a functor (see text below) ... ...Help will be appreciated ... ...

Peter=======================================================================================

*** NOTE ***

In order to help and answer the question posed in the above post, MHB readers of the post need access to Awodey's definition of a functor ... so I am providing access to the same ... as follows:View attachment 8345
https://www.physicsforums.com/attachments/8346Hope that helps ...

Peter

So you have a monoid-homomorphism $h:M \rightarrow N$ between the monoids $M$ and $N$.

The monoids $M$ and $N$ are viewed as categories, so they have both one object, say $\star$ and $\bullet$, respectively.

The functor $h$ must take objects to objects, so the only possibility is

$$h(\star) = \bullet$$

The functor $h$ must take arrows to arrows, so if $x:\star \rightarrow \star$ is an arrow in M, then $h(x)$ must be an arrow in N, that is $h(x):\bullet \rightarrow \bullet$. Thus we have

$$h(x):h(\star) \rightarrow h(\star)$$

The functor $h$ must take the unit of an object $x$ in $M$ to the unit of the object $h(x)$ in $N$.
$M$ has one object $\star$ and $N$ each has one object $\bullet$
The functor $h$ must take the unit of $\star$ to the unit of $h(\star)$.
The unit of the object of $M$ is $1_{\star}$, the unit of the object of $N$ is $1_{\bullet}$, Thus $h(1_{\star}) = 1_{\bullet}$, you see immediately that

$$h(1_{\star}) = 1_{h(\star)}$$

Can you now write down the composition rule ?

Last edited:
steenis said:
So you have a monoid-homomorphism $h:M \rightarrow N$ between the monoids $M$ and $N$.

The monoids $M$ and $N$ are viewed as categories, so they have both one object, say $\star$ and $\bullet$, respectively.

The functor $h$ must take objects to objects, so the only possibility is

$$h(\star) = \bullet$$

The functor $h$ must take arrows to arrows, so if $x:\star \rightarrow \star$ is an arrow in M, then $h(x)$ must be an arrow in N, that is $h(x):\bullet \rightarrow \bullet$. Thus we have

$$h(x):h(\star) \rightarrow h(\star)$$

The functor $h$ must take the unit of an object $x$ in $M$ to the unit of the object $h(x)$ in $N$.
$M$ has one object $\star$ and $N$ each has one object $\bullet$
The functor $h$ must take the unit of $\star$ to the unit of $h(\star)$.
The unit of the object $M$ is $1_{\star}$, the unit of the object $N$ is $1_{\bullet}$, Thus $h(1_{\star}) = 1_{\bullet}$, you see immediately that

$$h(1_{\star}) = 1_{h(\star)}$$

Can you now write down the composition rule ?
Thanks steenis ...

Your help is most welcome ...

I will be working on this tomorrow morning ...

I couldn't have progressed without your help ...

Thanks again ...

Peter

steenis said:
So you have a monoid-homomorphism $h:M \rightarrow N$ between the monoids $M$ and $N$.

The monoids $M$ and $N$ are viewed as categories, so they have both one object, say $\star$ and $\bullet$, respectively.

The functor $h$ must take objects to objects, so the only possibility is

$$h(\star) = \bullet$$

The functor $h$ must take arrows to arrows, so if $x:\star \rightarrow \star$ is an arrow in M, then $h(x)$ must be an arrow in N, that is $h(x):\bullet \rightarrow \bullet$. Thus we have

$$h(x):h(\star) \rightarrow h(\star)$$

The functor $h$ must take the unit of an object $x$ in $M$ to the unit of the object $h(x)$ in $N$.
$M$ has one object $\star$ and $N$ each has one object $\bullet$
The functor $h$ must take the unit of $\star$ to the unit of $h(\star)$.
The unit of the object of $M$ is $1_{\star}$, the unit of the object of $N$ is $1_{\bullet}$, Thus $h(1_{\star}) = 1_{\bullet}$, you see immediately that

$$h(1_{\star}) = 1_{h(\star)}$$

Can you now write down the composition rule ?
Thanks again for your help, Steenis ...

I will try to put your notation into the exact notation and form of Awodey's Definition 1.2 parts (a), (b) and (c) ... ... See below for Definition 1.2 ... ...Now Awodey says that a functor $$\displaystyle h : M \to N$$ between categories (in our case, monoids ... ) $$\displaystyle M$$ and $$\displaystyle N$$ is a mapping of objects to object and arrows to arrows in such a way that (a) , (b) and (c) of Definition 1.2 hold ... See below for Definition 1.2 ... ...

Consider the arrow $$\displaystyle x : \star \to \star$$ in $$\displaystyle M$$

then (a) becomes ...

$$\displaystyle h( x : \star \to \star ) = h(x) : h( \star ) \to h( \star )$$

$$\displaystyle \Longrightarrow h( x : \star \to \star ) = h(x) : \bullet \to \bullet$$and (b) becomes

$$\displaystyle h( 1_\star ) = 1_{ h( \star )}$$

$$\displaystyle \Longrightarrow h( 1_\star ) = 1_{ \bullet }$$ Then consider two arrows in $$\displaystyle M$$ ... namely $$\displaystyle x : \star \to \star$$ and $$\displaystyle y : \star \to \star$$ ... ...

so then (c) becomes

$$\displaystyle h( x \circ y ) = h(x) \circ h(y)$$ ... which is the composition rule ... ...Is the above correct?

Peter
=========================================================================================I believe it will be helpful to readers of the above post to have Awodey's Definition 1.2 easily available ... so I am again providing the definition ... as follows:View attachment 8349
https://www.physicsforums.com/attachments/8350
Hope that helps ...

Peter

Yes, correct, but with one remark

For two elements $x, y$ of monoid $M$ viewed as a monoid, the monoid homomorphism acts on the product as follows

$$h(x \cdot_M y)=h(x) \cdot_N h(y)$$

the product $x \cdot_M y$ in the monoid $M$ viewed as a monoid corresponds with the composition $x \circ y: \star \rightarrow \star$ in the monoid $M$ viewed as a category

The product $h(x) \cdot_N h(y)$ in the monoid $N$ viewed as a monoid corresponds with the composition $h(x) \circ h(y): \bullet \rightarrow \bullet$ in the monoid $N$ viewed as a category

So now we have

$$h(x \circ y)=h(x) \circ h(y)$$

The composition takes products to products
Peter said:
Observe that a monoid homomorphism from $$\displaystyle M$$ to $$\displaystyle N$$ is the same thing as a functor from $$\displaystyle M$$ regarded as a category to $$\displaystyle N$$ regarded as a category. ... ... "
Peter
Now you are halfway
Suppose you have monoids $M$ and $N$ are viewed as categories, they have both one object, say $\star$ and $\bullet$, respectively.
And there is a functor $H:M \rightarrow N$
This functor has the properties

action on the object of $M$: $H(\star)=\bullet$
action on an arrow $x:\star \rightarrow \star$ in $M$: $H(x)(\star) \rightarrow H(\star)$ in $N$
action on the unit $1_\star$ in $M$: $H(1_\star)=1_{H(\star)}$, which is the unit $1_\bullet$ in $N$

and the composition rule for two arrows $x, y$ in $M$: $H(x \circ y)=H(x) \circ H(y)$, which is the composition of two arrows $H(x), H(y)$ in $N$Can you show that this functor $H$ between the monoids $M$ and $N$ viewed as categories is also a monoid homomorphism between the monoids $M$ and $N$ viewed as monoids ?

steenis said:
Yes, correct, but with one remark

For two elements $x, y$ of monoid $M$ viewed as a monoid, the monoid homomorphism acts on the product as follows

$$h(x \cdot_M y)=h(x) \cdot_N h(y)$$

the product $x \cdot_M y$ in the monoid $M$ viewed as a monoid corresponds with the composition $x \circ y: \star \rightarrow \star$ in the monoid $M$ viewed as a category

The product $h(x) \cdot_N h(y)$ in the monoid $N$ viewed as a monoid corresponds with the composition $h(x) \circ h(y): \bullet \rightarrow \bullet$ in the monoid $N$ viewed as a category

So now we have

$$h(x \circ y)=h(x) \circ h(y)$$

The composition takes products to productsNow you are halfway
Suppose you have monoids $M$ and $N$ are viewed as categories, they have both one object, say $\star$ and $\bullet$, respectively.
And there is a functor $H:M \rightarrow N$
This functor has the properties

action on the object of $M$: $H(\star)=\bullet$
action on an arrow $x:\star \rightarrow \star$ in $M$: $H(x)(\star) \rightarrow H(\star)$ in $N$
action on the unit $1_\star$ in $M$: $H(1_\star)=1_{H(\star)}$, which is the unit $1_\bullet$ in $N$

and the composition rule for two arrows $x, y$ in $M$: $H(x \circ y)=H(x) \circ H(y)$, which is the composition of two arrows $H(x), H(y)$ in $N$Can you show that this functor $H$ between the monoids $M$ and $N$ viewed as categories is also a monoid homomorphism between the monoids $M$ and $N$ viewed as monoids ?
Thanks Steenis ... most helpful ...

Reflecting on your post now ...

Peter

steenis said:
Yes, correct, but with one remark

For two elements $x, y$ of monoid $M$ viewed as a monoid, the monoid homomorphism acts on the product as follows

$$h(x \cdot_M y)=h(x) \cdot_N h(y)$$

the product $x \cdot_M y$ in the monoid $M$ viewed as a monoid corresponds with the composition $x \circ y: \star \rightarrow \star$ in the monoid $M$ viewed as a category

The product $h(x) \cdot_N h(y)$ in the monoid $N$ viewed as a monoid corresponds with the composition $h(x) \circ h(y): \bullet \rightarrow \bullet$ in the monoid $N$ viewed as a category

So now we have

$$h(x \circ y)=h(x) \circ h(y)$$

The composition takes products to productsNow you are halfway
Suppose you have monoids $M$ and $N$ are viewed as categories, they have both one object, say $\star$ and $\bullet$, respectively.
And there is a functor $H:M \rightarrow N$
This functor has the properties

action on the object of $M$: $H(\star)=\bullet$
action on an arrow $x:\star \rightarrow \star$ in $M$: $H(x)(\star) \rightarrow H(\star)$ in $N$
action on the unit $1_\star$ in $M$: $H(1_\star)=1_{H(\star)}$, which is the unit $1_\bullet$ in $N$

and the composition rule for two arrows $x, y$ in $M$: $H(x \circ y)=H(x) \circ H(y)$, which is the composition of two arrows $H(x), H(y)$ in $N$Can you show that this functor $H$ between the monoids $M$ and $N$ viewed as categories is also a monoid homomorphism between the monoids $M$ and $N$ viewed as monoids ?

Thanks again Steenis ...

Now ... you write:

" ... ... Can you show that this functor $H$ between the monoids $M$ and $N$ viewed as categories is also a monoid homomorphism between the monoids $M$ and $N$ viewed as monoids? ... ... "The essential point is that the product in the monoid concerned corresponds to the composition of arrows in the category ...

So ... $$\displaystyle H( x \circ y ) \equiv H( x \bullet_M y )$$ where $$\displaystyle \bullet_M$$ is multiplication in the monoid $$\displaystyle M$$

and ... $$\displaystyle H(x) \circ H(y) \equiv H(x) \bullet_N H(y)$$

and so given the above ... we have

$$\displaystyle H( x \bullet_M y ) = H(x) \bullet_N H(y)$$

and of course, we also have ...

$$\displaystyle H( 1_\star ) = 1_\bullet$$
Is the above correct ...

Peter

OK, Peter, correct.

steenis said:
OK, Peter, correct.
Thanks for all your help, Steenis ...

Peter

## 1. What is a monoid?

A monoid is a mathematical structure that consists of a set of elements and a binary operation that is associative and has an identity element. In simpler terms, it is a set with a single operation that follows specific rules.

## 2. How are monoids related to categories?

Monoids can be thought of as a special type of category, where the objects are the elements of the monoid and the morphisms are the operation. This allows us to use the concepts and tools from category theory to better understand monoids.

## 3. What is the significance of Example 13 in Awodey Section 1.4?

Example 13 in Awodey Section 1.4 illustrates the concept of a monoid as a category in a concrete and relatable way. It shows how the set of natural numbers with addition as the operation forms a monoid and how this monoid can be viewed as a category with a single object and multiple morphisms.

## 4. Can you provide another example of a monoid as a category?

Yes, another example of a monoid as a category is the set of strings with concatenation as the operation. The set of strings can be thought of as a category with each string as an object and the concatenation of strings as the morphism between them.

## 5. What is the benefit of understanding monoids as categories?

Understanding monoids as categories allows us to make connections and draw comparisons between different mathematical structures. It also provides a way to use the tools and concepts from category theory to study and analyze monoids, ultimately leading to a deeper understanding of both monoids and category theory.

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