Understanding Categories: A Question on Commutativity

[topsquark]

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

 

[Evgeny.Makarov]

\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.

Given f and g I can always find an \(\displaystyle \alpha\) and \(\displaystyle \beta\).

This is not obvious to me.

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...".

Do \(\displaystyle \alpha\) and \(\displaystyle \beta\) equal m and n respectively

It is not clear why this should always be so.

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.​

 

[topsquark]

\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.

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.

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.

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.

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​

 

[Evgeny.Makarov]

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 $\alpha:A\to C$.

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 $C$.

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.

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.

 

[topsquark]

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

 

[Joppy]

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?

 

[Evgeny.Makarov]

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.