Understanding Monoids as Categories: Awodey Section 1.4 Example 13

[Math Amateur]

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

 

[steenis]

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 ?

 

[Math Amateur]

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

 

[Math Amateur]

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

 

[steenis]

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

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 ?

 

[Math Amateur]

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

 

[Math Amateur]

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

 

[steenis]

OK, Peter, correct.

 

[Math Amateur]

OK, Peter, correct.

Thanks for all your help, Steenis ...

Peter