# Understanding Categories: A Question on Commutativity

• MHB
• topsquark
In summary: I'll have to think about this some more.In summary, the conversation discusses the concept of categories and their objects and morphisms. The problem statement introduces the category \mathcal{D}, whose objects are all the morphisms of \mathcal{C}. The conversation then explores the relationship between morphisms in \mathcal{D} and whether they are equal to other morphisms in the category. It also raises questions about the existence and generalizability of morphisms between certain objects in a category.
topsquark
Gold Member
MHB
I just started to upgrade my skills by learning a little about categories.

There was an example in my text that I'd like to get some feedback on.

Let $$\displaystyle \mathcal{C}$$ be any category and define the category $$\displaystyle \mathcal{D}$$ whose objects are all morphisms of $$\displaystyle \mathcal{C}$$. If $$\displaystyle f: A \to B$$ and $$\displaystyle g: C \to D$$ are morphisms of $$\displaystyle \mathcal{C}$$, then hom(f, g) consists of all pairs $$\displaystyle \alpha, ~ \beta$$, where $$\displaystyle \alpha : A \to C$$ and $$\displaystyle \beta : B \to D$$ are morphisms of $$\displaystyle \mathcal{C}$$ such that the following diagram is commutative.
(You have some pretty software on the site that would make the diagram nice. Unfortunately I haven't learned to us it so just use your imagination...)
$$\displaystyle A ~ \begin{matrix} f \\ \longrightarrow \end{matrix} ~ B \begin{matrix} \beta \\ \longrightarrow \end{matrix} ~ D$$

$$\displaystyle A ~ \begin{matrix} \alpha \\ \longrightarrow \end{matrix} ~ C \begin{matrix} g \\ \longrightarrow \end{matrix} ~ D$$

where the diagram is the usual rectangular one.

Okay, to my question. Given f and g I can always find an $$\displaystyle \alpha$$ and $$\displaystyle \beta$$. But how general are we being here? For example, $$\displaystyle \mathcal{D}$$ also contains morphisms $$\displaystyle m: A \to C$$ and $$\displaystyle n: B \to D$$. Do $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ equal m and n respectively or is it that $$\displaystyle \alpha$$ is an element of m and $$\displaystyle \beta$$ is an element of n chosen so the diagram will be commutative?

I worked out a few simple examples and was able to construct $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ each time but I couldn't find a way to say that any morphisms from the sets m and n wouldn't work. I can't find a proof of the situation either way.

Thanks!

-Dan

\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
%preamble \usetikzlibrary{positioning}
\node (A) {$A$};
\node[right=of A] (B) {$B$};
\node[below=of A] (C) {$C$};
\node[right=of C] (D) {$D$};
\draw[->] (A) -- node[above] {$f$} (B);
\draw[->] (C) -- node[below] {$g$} (D);
\draw[->] (A) -- node
{$\alpha$} (C);
\draw[->] (B) -- node
{$\beta$} (D);
\end{tikzpicture}
I am not a specialist in category theory, but here are my two cents.

topsquark said:
Given f and g I can always find an $$\displaystyle \alpha$$ and $$\displaystyle \beta$$.
This is not obvious to me.

topsquark said:
For example, $$\displaystyle \mathcal{D}$$ also contains morphisms $$\displaystyle m: A \to C$$ and $$\displaystyle n: B \to D$$.
You probably mean, "Suppose $$\displaystyle \mathcal{D}$$ also contains...".

topsquark said:
Do $$\displaystyle \alpha$$ and $$\displaystyle \beta$$ equal m and n respectively
It is not clear why this should always be so.

topsquark said:
or is it that $$\displaystyle \alpha$$ is an element of m and $$\displaystyle \beta$$ is an element of n chosen so the diagram will be commutative?
The relation "is an element of" is not defined on morphisms in general.​

Evgeny.Makarov said:
\begin{tikzpicture}[>=stealth']
%preamble \usetikzlibrary{arrows}
%preamble \usetikzlibrary{positioning}
\node (A) {$A$};
\node[right=of A] (B) {$B$};
\node[below=of A] (C) {$C$};
\node[right=of C] (D) {$D$};
\draw[->] (A) -- node[above] {$f$} (B);
\draw[->] (C) -- node[below] {$g$} (D);
\draw[->] (A) -- node
{$\alpha$} (C);
\draw[->] (B) -- node
{$\beta$} (D);
\end{tikzpicture}
I am not a specialist in category theory, but here are my two cents.

Thanks for getting back to me. I appreciate it.

Evgeny.Makarov said:
This is not obvious to me.
I should mention that the only examples I looked at were all finite sets. I may have not been using a big enough bush.

Evgeny.Makarov said:
You probably mean, "Suppose $$\displaystyle \mathcal{D}$$ also contains...".
The problem statement says "category D whose objects are all morphisms of C" I was taking that to mean that all the possible morphisms of $$\displaystyle \mathcal{C}$$ are in $$\displaystyle \mathcal{D}$$. Perhaps that isn't correct.

Evgeny.Makarov said:
It is not clear why this should always be so.
Which is one of the nagging questions I have about this. I don't see if they would have to be either. I can't think of a way to prove or disprove this statement. Even the examples I work out didn't imply anything one way or another.

Evgeny.Makarov said:
The relation "is an element of" is not defined on morphisms in general.
Hmmmm... Now that I am looking for it the text doesn't use this term either. But they still refer to "objects" that are "in" a category. What language is used for categories?

Thanks again!

-Dan​

topsquark said:
Given $f$ and $g$ I can always find an $\alpha$ and $\beta$.
A category does not have to have morphisms between every two objects. For example, in the category of sets there is no morphisms from a nonempty set to the empty one. So if $A\ne\emptyset$ and $C=\emptyset$ there is no $\alpha:A\to C$.

topsquark said:
The problem statement says "category D whose objects are all morphisms of C" I was taking that to mean that all the possible morphisms of $$\displaystyle \mathcal{C}$$ are in $$\displaystyle \mathcal{D}$$.
.Yes, $\mathcal{D}$ contains all morphisms of $\mathcal{C}$, but the previous example shows that there may be no morphisms between $A$ and $C$.

topsquark said:
Do $\alpha$ and $\beta$ equal $m$ and $n$ respectively?
Not necessarily. For example, suppose $B=\{b\}$ and $g(x)=\beta(b)$ for all $x\in C$. Then any function from $A$ to $C$ makes the diagram commute.

topsquark said:
Now that I am looking for it the text doesn't use this term either. But they still refer to "objects" that are "in" a category. What language is used for categories?
Yes, objects can be in a category. But you were saying "$\alpha$ is an element of $m$" where $\alpha$ and $m$ are morphisms of $\mathcal{C}$, if I understand correctly. And "is an element of" is not defined on morphisms.

Evgeny.Makarov said:
A category does not have to have morphisms between every two objects. For example, in the category of sets there is no morphisms from a nonempty set to the empty one. So if $A\ne\emptyset$ and $C=\emptyset$ there is no $\alpha:A\to C$.

.Yes, $\mathcal{D}$ contains all morphisms of $\mathcal{C}$, but the previous example shows that there may be no morphisms between $A$ and $C$.

Not necessarily. For example, suppose $B=\{b\}$ and $g(x)=\beta(b)$ for all $x\in C$. Then any function from $A$ to $C$ makes the diagram commute.

Yes, objects can be in a category. But you were saying "$\alpha$ is an element of $m$" where $\alpha$ and $m$ are morphisms of $\mathcal{C}$, if I understand correctly. And "is an element of" is not defined on morphisms.
Got it. Thanks for the feedback. It's a lot clearer now.

-Dan

Evgeny.Makarov said:
Yes, objects can be in a category. But you were saying "$\alpha$ is an element of $m$" where $\alpha$ and $m$ are morphisms of $\mathcal{C}$, if I understand correctly. And "is an element of" is not defined on morphisms.

So the language for talking about an object belonging to a category is to state it in plain English?

Formally a category is a triple whose first element is a class of objects. A class, because, strictly speaking, it may not be a set, for example, when objects are all sets. So an unambiguous phrase is "$A$ belongs to the class of objects of category $\mathcal{C}$". I am sure "$A$ is an object of $\mathcal{C}$" is OK. I also don't think saying "Object $A$ belongs to category $\mathcal{C}$" causes any confusion.

## 1. What is commutativity?

Commutativity is a mathematical property that refers to the ability to change the order of operations or elements in a mathematical equation without changing the final result.

## 2. Why is understanding commutativity important?

Understanding commutativity is important because it allows us to simplify and solve mathematical problems more efficiently. It also helps us to better understand the relationships between different mathematical operations and their properties.

## 3. What are some examples of commutative operations?

Addition and multiplication are the most common examples of commutative operations. For example, 3 + 5 is equal to 5 + 3, and 2 x 4 is equal to 4 x 2.

## 4. Are all mathematical operations commutative?

No, not all mathematical operations are commutative. Subtraction and division are examples of non-commutative operations. For example, 6 - 3 is not equal to 3 - 6, and 12 ÷ 4 is not equal to 4 ÷ 12.

## 5. How does commutativity relate to other mathematical properties?

Commutativity is closely related to other mathematical properties, such as associativity and distributivity. These properties all describe how operations can be rearranged to achieve the same result. Commutativity specifically refers to the ability to switch the order of operations or elements, while associativity refers to the ability to group operations differently, and distributivity refers to the ability to distribute an operation over multiple terms.

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