Abstract Nonsense

Crushing one theorem at a time

Universal Arrows and Universal Elements (Pt. II)


Point of Post: This is a continuation of this post.

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To get a richer class of examples we consider a more specific type of universal arrow. Suppose that we had some functor F:\mathcal{C}\to\mathbf{Set} and some singleton set \ast. What would a universal arrow from \ast to F look like? Well, such a universal arrow would consist of \mathcal{C}-object c and an arrow \ast\to F(c) such that given any other \mathcal{C}-object x with a specified arrow \ast\to F(x) there exists a unique arrow c\xrightarrow{j}x whose image under F commutatively completes the triangle given by F(c)\longleftarrow\ast\longrightarrow F(x). Ok, fine, we literally just wrote down the definition of a universal arrow from \ast to F, so how can we make things simpler? The key observation is recognizing (really) what an arrow \ast\to X (where X is any set) really is. Truly, yes, it’s a function, but for all practical purposes we can think about such an arrow as “picking out” some element x_0 of X (the one corresponding to the image of the arrow). So, let’s recast our discussion in this language. We see then what we are really when we are talking about a universal arrow from \ast to F is some \mathcal{C}-element c along with some distinguished element a_0\in F(c) with the property that given any other \mathcal{C}-element x and a distinguished element b_0\in F(x) there exists a unique arrow c\xrightarrow{j}x such that F(j) respects the distinguished objects (i.e. F(j)(a_0)=b_0). Such a pair (c,a_0) is called a universal element for F.

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There are two classic examples of universal elements (both of which can be found in [1]) which I can’t help but discuss.

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Fix some group G and some normal subgroup N\unlhd G. Suppose that we had another group, H and some homomorphism f:G\to H. A natural question that comes up is when can we factor f through the canonical projection \pi:G\to G/N? In other words, under which conditions can we find some homomorphism g:G/N\to H such that f=g\circ\pi? Well, this question is quickly answered in the first homomorphism theorem, a ‘lemma’ (really the whole proof, or the important parts of it) to the first isomorphism theorem, which says that such a factorization is possible precisely when N\subseteq\ker f. But, this tells us that \pi is a universal element of a certain functor. Indeed, let F be the functor \mathbf{Grp}\to\mathbf{Set} given by sending H to

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F(H)=\left\{f\in\text{Hom}_\mathbf{Grp}(G,H):N\subseteq\ker f\right\}

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and sending g:G\to H to g^\ast where, as in the case of the covariant Hom functor, we define g^\ast(h)=g\circ h if h:H\to H'. We claim that (G/N,\pi) is a universal element for F. Indeed, suppose we are given some group H and some distinguished f\in F(H). Since f annihilates N we have by the first homomorphism theorem that f=g\circ \pi for some g:G/N\to H. Thus, g is an arrow G/N\to H and F(g)(\pi)=g\circ\pi=f. Moreover, it’s clear that g is unique, since \pi is epic (being surjective). This proves that (G/N,\pi) is a universal element for F. The cool thing is that (as we have remarked before) this is really the characterization of being a quotient group–it’s what really matters. In particular, all quotient group related theorems are recoverable from the fact that (G/N,\pi) is a universal element for F. For example, the clever proof of the first isomorphism theorem shows that if f:G\to H is a surjective group homomorphism (an epimorphism in \mathbf{Grp}) with kernel N then H\cong G/N by (implicitly) showing that (H,f) is a universal element for F showing that H and G/N must be isomorphic (via a unique isomorphism!).

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The other classic example is something we have recently been discussing. Namely, fix some commutative ring R and consider two left R-modules M and N. We can then consider the set-valued functor B:R\text{-}\mathbf{Mod}\to\mathbf{Set} given by B(L)=\text{Bil}(M,N;L) on objects (where, as we have previously discussed, \text{Bil}(M,N;L) is the set of bilinear maps M\times N\to L) and given an arrow g:L\to L' we define B(g)=g^\ast where (as usual) g^\ast(h)=g\circ h. So, what does a universal element for B look like? Well, if you haven’t guessed it by now (hard to guess, yes I know) a universal element is given by (M\otimes_R N,\otimes) where M\otimes_R N is the tensor product and \otimes is the tensor map. Why is this true? Well, probably because it’s how we (implicitly) defined the tensor product! Indeed, we defined the tensor product to be an initial object in the comma category (\ast\downarrow B), but by the above discussion this is the same thing as defining the tensor product to be a universal arrow from \ast to B, or as we are now calling such an object, a universal element for B. Neat, huh?

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As a last remark we define the “dual” concept to a universal arrow, a couniversal arrow. More descriptively we can think of couniveral arrows as being universal arrows away from a given functor. To be more specific, if we have some functor F:\mathcal{C}\to\mathcal{D} and some object d of \mathcal{D} we define a couniversal arrow from d to F or a universal arrow F to d to be an ordered pair (c,u) where c is some \mathcal{C}-object and u is an arrow F(c)\to d such that given any other \mathcal{C}-object x and an arrow F(x)\xrightarrow{f}d there exists a unique arrow x\xrightarrow{j}c such that u\circ F(j)=f.

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References:

[1] Mac, Lane Saunders. Categories for the Working Mathematician. New York: Springer-Verlag, 1994. Print.

[2] Adámek, Jirí, Horst Herrlich, and George E. Strecker. Abstract and Concrete Categories: the Joy of Cats. New York: John Wiley & Sons, 1990. Print.

[3] Berrick, A. J., and M. E. Keating. Categories and Modules with K-theory in View. Cambridge, UK: Cambridge UP, 2000. Print.

[4] Freyd, Peter J. Abelian Categories. New York: Harper & Row, 1964. Print.

[5] Rotman, Joseph J. Introduction to Homological Algebra. Springer-Verlag. Print.

[6] Herrlich, Horst, and George E. Strecker. Category Theory: An Introduction. Lemgo: Heldermann, 2007. Print.

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January 12, 2012 - Posted by | Algebra, Category Theory | , , , , ,

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