- #1

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

(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...)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.

\(\displaystyle A ~ \begin{matrix} f \\ \longrightarrow \end{matrix} ~ B \begin{matrix} \beta \\ \longrightarrow \end{matrix} ~ D\)

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

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