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I am trying to learn some category theory and have noticed that sometimes parallel arrows appear in certain diagrams (for example: the definition of equalizer and coequalizer).

It seems to me that the usual intepretation of $$X\overset{f}\rightarrow Y \overset{g}{\underset{h}\rightrightarrows} Z$$ is $$\begin{matrix} X&\overset{f}\rightarrow&Y\\ \;\;\downarrow_f&&\;\;\downarrow_g\\ Y&\overset{h}\rightarrow&Z \end{matrix}$$

and the usual interpretation of the diagram $$X\overset{g}{\underset{h}\rightrightarrows} Y \overset{f}\rightarrow Z$$ is $$\begin{matrix} X&\overset{g}\rightarrow&Y\\ \;\;\downarrow_h&&\;\;\downarrow_f\\ Y&\overset{f}\rightarrow&Z \end{matrix}$$

At least with these two interpretations it seems that the definitions of equalizers and coequalizers can be phrased nicely in terms of commutative squares.

How do we interpret parallel arrows in general? For example, how does one interpret something like $X\overset{f}\rightarrow Y \overset{g}{\underset{h}\rightrightarrows} Z\overset{k}\rightarrow W$?

I guess what I'm looking for is a general interpretation of parallel arrows in terms of commutative squares, if that makes sense.

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It's not clear to me why you think the square view is clearer than the original view.

As you get more advanced in category theory, you'll see that equalizers and co-equalizers are part of a general view of "limits" and "co-limits," and then, after that, you encounter the notion of "adjoint functors," which is a wonderful catch-all that describes a huge number of types of operations.

In general, if $G$ is a directed multi-graph, then there is an associated small category, $\mathcal D_G$, having the nodes of $G$ as objects and the maps from $x$ to $y$ being directed paths from $x$ to $y$. (In most cases we find interesting, $G$ is acyclic, too, but nothing requires that.)

A $G$-diagram in a category $\mathcal C$ is just a functor $F:\mathcal D_G\to\mathcal C$.

A $G$-diagram "commutes" if, for all $x,y\in G$ and $p,q\in Hom_{\mathcal D_G}(x,y)$, $F(p)=F(q)$. This means, essentially, that any path in the graph $G$ corresponds to the same composed function in $\mathcal C$.

The collection of all commuting $G$-diagrams in $\mathcal C$ is, itself, a category, and there is a trivial functor from $\mathcal C$ to the category of commuting $G$-diagrams by creating constant $G$-diagrams.

The equalizer and co-equalizer, when they exist, turn out to be something called "adjoints" of this functor. Alternatively, you can think of them as "limits" and "colimits" of $G$-diagrams, where $G$ is a graph with two nodes $x,y$ and two distinct arrows $x\to y$.

If $G$ is a graph with two nodes and no edges, then the "limit and co-limit" of a $G$-diagrams are just the category "sum" and "product," for example.

The square diagram is harder to put into this more general frame. You essentially end up with the above view, but with more contortions in the definition of your category.

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What about drawing up both squares?

Or a hexagon: the upper 3 arrows go to the right: $f,g,k$, and below, again to the right $f,h,k$. And, I assume, you want the hexagon to commute.

(Well, these 2 interpretations don't mean the same thing: drawing both squares mean more: $fg=fh$ and $gk=hk$. Which interpretation you intend to use?)

However, in general, unless it is said that the given diagram involving the parallel arrows is a (co-)equalizer, we usually don't impose any condition to parallel arrows $X\overset{g}{\underset{h}\rightrightarrows} Y$.

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Right! So, if I understand you correctly, the proper way to interpret two parallel arrows would be like this: $$\begin{matrix} X&\overset{f}\rightarrow&Y\\ &&\\ X&\overset{g}\rightarrow&Y \end{matrix}$$ i.e. discretely. Interesting. So the equalizer (when defined using parallel arrows) is actually defined using two diagrams? – Dejan Govc Nov 1 '12 at 13:01
$X\overset{g}{\underset{h}\rightrightarrows} Y$ is just visualization of $g,h\in\operatorname{Hom}(X,Y)$. There isn't really much interpreting necessary. – roman Nov 1 '12 at 13:10
@Berci: I apologize, it seems I was a bit hasty in accepting your answer. Thank you very much for your helpful thoughts. – Dejan Govc Nov 1 '12 at 14:18

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