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I'm having trouble understanding the definition of a universal arrow in Maclane's CWM. In particular, what are the fixed arguments?
I know at least the functor is fixed, but I can't tell if $c$ is also.
And, by the way, what good are universal arrows for? What do they tell us?
Anything will help.
I shall use CWM's notation very closely. We will use an example "as simple as possible , but not any simpler", as Einstein used to say.
Let $\mathcal C$ and $\mathcal D$ be the following categories
Where $f_3$ is $f_{12} \circ f_1$
For the diagrams, please see comment below.
Define a functor $S$ between the two categories: $S(d_1)=c_1$, $S(d_2)=c_2$, $S(h)=f_{12}$
We can display the 2 categories side by side as CWM does. Note that the domain category $\mathcal D$ is on the right. More on this later.
Let's consider CWM's definition. Fix $c$ and $S$. (this answers your "what is fixed" question). CWM's c will be our $c_0$. Let's thus see if it is true or not that there is a universal arrow from $c_0$ to $S$. Unfold the definition: "for all pairs $<d,f:c_0 \to S(d)>$ can we factor $f$ as....?". There are only 2 such pairs: $<d_1, f_1>$ and $<d_2, f_3>$. Indeed, we see that $f_1=S(Id_{d_1}) \circ f_1$ and $f_3=S(h) \circ f_1$ . So$f_1$ acts as a universal gateway $u$ which allows us to go to any S-object $(S(d_1), S(d_2))$ in a unique way (different in each case) with the help of S-morphisms $(S(Id_{d_1}), S(h))$. From the above factorizations, we see that to each $f:c_0 \to S(d)$ we can associate a $f'$ in $\mathcal D$ as follows:$f_1' = Id_{d_1}$, $f_3' = h$. There are no other $f'$ s. Conclusion: a universal arrow from $c_0$ to $S$ exists and it is $<r,u> = <d_1, f_1>$. Now rerun the CWM's definition with $c= c_1$. Do we have a universal arrow in this case? Yes!, but this time is $<r,u>=<d_1, Id_{d_1}>$. What if we try with $c=c_2$? Again yes! with universal arrrow $<r,u>=<d_2, Id_{d_2}>$. So we see that universal arrows depend very much on the chosen $c$. Therefore this functor $S$ has the following remarkable property: for all $c$ in $\mathcal C$, there is a universal arrow from $c$ to $S$. As is shown in CWM's chapter on Adjoints, this guarantees us that $S$ is a right adjoint functor and that a left adjoint functor $F$ exists. The object function of $F$ is easy to find: we saw that to each $c$ we can associate a "universal object" $r$ in $\mathcal D$. So we set $r= F (c)$. Specifically: $F(c_0)=d_1$, $F(c_1)=d_1$,$F(c_2)=d_2$. The arrow function of $F$ is also easy to find in this case (not many choices) and is: $F(f_1)= Id_{d_1}$, $F(f_{12})=h$, $F(f_3)=h$ and it indeed respects the functor composition rules.
In CWM's definition of adjunction, you will see a diagram. CWM's $X$ is our $\mathcal C$, $A$ is our $\mathcal D$, $F$ is our $F$, and $G$ is our $S$. Now you see why CWM's and my diagrams above show $\mathcal D$ on the right side. That is because these $S$ functors are often/hopefully right adjoints.
An example of a functor without a universal arrow? Easy: take category $\mathcal D'$ like $\mathcal D$ but with an arrow $h'$ parallel to $h$ and functor $S'$ like $S$ but also with $S'(h')=f_{12}$. Now: $S'$ does not have a universal arrow from $c_0$ to $S'$, since the unicity requirement of $f_3'$ is not met.
You may also look at the definition of universal arrow as the initial object in the comma category of "objects S-under c". Modern terminology calls this a slice category. It is very simple in this case
$<d_1,f_1>$ is indeed the initial object in the slice category $<c\downarrow S>$, so it is a universal arrow from $c$ to $S$.
Now a medieval metaphor to "visualize" universal arrows: imagine the S-image is a castle. $c$ is a point outside the castle. If there is any way for you to go from $c$ outside to a point $S(d)$ inside, then surely you must go first to the main gate $S(r)$ via path $u$, then go to $S(d)$ via a unique path $S(f')$ inside the castle.
I don't have Cat for the Working Mat with me, but wiki has a good explanation about the definition of universal arrows.
As per why they are useful, they are useful because they allow us to discuss universal characterizations. These are arrow theoretic characterizations of objects in a category that specify them up to unique isomorphism. This idea is perhaps one of the most powerful notions in the entirety of category theory. We live in a very categorical, arrow-obsessed world. Most of the modern definitions of things involve and become simpler when we are able to phrase them entirely in terms of category theory. In fact, most proofs become trivial when viewed in this purely arrow-theoretically defined way.
Let me illustrate:
Suppose that you have an abelian group $A$, a collection of abelian groups $B_n$, and $\displaystyle B=\prod_n B_n$. It is a theorem, probably known to you, that as groups
$$\text{Hom}\left(A,B\right)=\prod_n\text{Hom}(A,B_n)$$
Now, if you defined $\displaystyle \prod_n B_n$ to be "the set of all tuples..." then this is non-obvious. But, one could have used the following universal characterization of a product:
Let $X_n$ be a sequence of abelian groups. A product of the $X_n$ is a group $X$ and a set of group maps $p_n:X\to X_n$ such that given any abelian group $A$ and a set of group maps $f_n:A\to X_n$ there exists a unique group map $f:A\to X$ such that $p_n\circ f=f_n$
You can quickly check that our usual definition of $X=\displaystyle \prod_n X_n$ along with the canonical projection maps $\text{proj}_n:X\to X_n$ satisfy these definitions. Then, this theorem really just becomes a tautological restatement of the definition of a product!
Yes, you may argue that I have pulled the old French trick of turning a theorem into a definition, but let me give you another example.
I assume you are aware of the tensor product $M\otimes_R N$ of two $R$-modules ($R$ is commutative and unital for convenience sake). I doubt that you could find any modern user of mathematics that would tell you that the tensor product is not an indispensable tool.
So, what is the tensor product? Well, as should be obvious, one merely takes the free $R$-module on $M\times N$ and quotients out by the submodule generated by the relations $a(m,n)-(am,n)$, $a(m,n)-(m,an)$, $(m+n,l)-(m,l)-(n,l)$....
I doubt that any algebraist worth their salt thinks about tensor products this way--it's just a garbled mess. The way that one should think about tensor products is by the following universal characterization:
Let $M$ and $N$ be $R$-modules. Then, an $R$-module $T$ along with an $R$-bilinear map $b:M\times N\to T$ is a tensor product of $M$ and $N$ if for any $R$-module $L$, and any $R$-bilinear map $f:M\times N \to L$ there exists a unique $R$-module map $g:T\to L$ such that $f=g\circ b$.
This horribly mentioned construction then becomes necessary to mention, only once!, so that we know tensor products always exist. You see, the operative part about a tensor product's definition is that they allow you to trade in bilinear maps for linear maps--they turn multilinear algebra into linear algebra.
So, in the extremely important example of tensor products, one sees that universal constructions aren't just a convenient rephrasing of a definition but the only sensible one
