Assume the left square is a pushout. Then the right square is a pushout if and only if the big rectangle is.

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1 COMMUTATIVE ALGERA LECTURE 2: MORE CATEGORY THEORY VIVEK SHENDE Last time we learned about Yoneda s lemma, and various universal constructions initial and final objects, products and coproducts (which turned out to be the same for R-modules); pullbacks and pushouts. We connected this back to the usual discussion of exact sequences, by noting that if you turn into 0 A C 0 A 0 C then the sequence is left-exact if this diagram is a pullback; right exact if this diagram is a pushout, and exact if it s both. We left off by noting that the first isomorphism theorem says that whenever A, there s a pushout-pullback diagram A 0 /A The real content is in the assertion that it s a pushout (i.e. right exact); once you know this, the fact that it s a pullback (i.e. left exact) is already contained in the hypothesis that A. Here s a really important fact about pushouts. (There s a similar one about pullbacks.) Lemma 1. Suppose given a diagram like this: Assume the left square is a pushout. Then the right square is a pushout if and only if the big rectangle is. Proof. Exercise. 1

2 2 VIVEK SHENDE The second isomorphism theorem is about when you have submodules A, C. Note in in this case the following diagram is a (pullback and) pushout. A A A + Now apply the above lemma to the following diagram: A A 0 A + It tells you that the same item makes both the right square and the big rectangle pushouts. Thus you learn that /(A ) = (A + )/A. The third isomorphism theorem is about the situation A C. Thus consider the following diagram A C 0 /A C/A 0 Applying the lemma, we learn that the upper right square is a pushout. Applying it again, we learn makes the right rectangle and lower right square pushouts. I.e., C/ = (C/A)/(/A). As you see, the first isomorphism theorem is about one square, the second is about two squares, and the third is about three squares. That s why they re called the first, second, and third isomorphism theorems. 1 Now we return to category theory, to learn about adjoints. This has to do with a situation when you have two categories, C, D, and functors between them, f : C D : g In this situation, we say that an adjunction, with f as the left adjoint and g as the right adjoint, is a natural equivalence 1 Not really. Hom D (f(c), d) = Hom C (c, g(d))

3 COMMUTATIVE ALGERA LECTURE 2: MORE CATEGORY THEORY 3 That is, both sides define functors C op D set, and we demand the data of natural transformations between these functors, which compose both ways to the identity. Exercise: using Yoneda s lemma or otherwise, show that an adjoint, if it exists, is unique up to unique natural transformation. The prototypical example of an adjunction is (free construction, forgetful functor). E.g, we have a functor Rings Sets, which forgets the ring structure and just remembers the elements of the ring. It s a forgetful functor. On the other hand, given a set S, we can form the polynomial ring Z[S], where the variables are elements of S. There s a natural isomorphism Hom Rings (Z[S], R) = Hom Sets (S, F orget(r)) Actually, being the left adjoint of the forgetful functor is what it means to be a free construction. You should regard the above isomorphism as the assertion that there s a such thing as a free ring on a set S, and it s Z[S]. Similarly, you can forget that a module is a module, and remember only that it s a set. Given a set S, you can form the R-module with basis S, I ll denote it R S. This is the free R-module on S, i.e. there s an adjunction Hom R mod (R S, M) = Hom Sets (S, F orget(m)) The most useful consequence of the existence of an adjunction is: Lemma 2. Left adjoints preserve colimits and right adjoints preserve limits. We ll prove this shortly, but first let s review (or learn) what limits and colimits are. You ve seen many universal constructions, described in the following way. There s a diagram of objects and morphisms, and you ask for an object with morphisms (from) to all objects in the diagram, commuting with all morphisms in the diagram, which is (co) universal in the sense that any such object and collection of maps will factor through the a map (from) to the (co) universal one. E.g. given X W Y, the pullback is some X W Y, such that there s a commutative diagram X W Y Y X such that any other commutative diagram W Z Y X W is obtained from it by factoring through some unique map Z X W Y.

4 4 VIVEK SHENDE The categorical term for (co)universal constructions of this form is (co) limit. Here are some you have seen. limit final object product pullback kernel colimit initial object coproduct pushout cokernel Example. The forgetful functor from rings to sets has a left adjoint, so according to Lemma 2, it preserves limits. Here s what you learn from this: whatever the final object is in rings, its underlying set is the final object of sets, i.e. has one element. Whatever the product is in rings, its underlying set has to be the cartesian product of sets. You get a similar description for the underlying set of a pullback. In other words, if you first encountered the category of rings and wanted to know whether it had a final object, products, etc., then it is quite helpful to know that the forgetful functor has a left adjoint (i.e. that there s a free construction), so you know where to start looking for products. Conversely, we know that the initial object in rings in Z. The underlying set not being the empty set, we see that the forgetful functor can t have a right adjoint. Example. The forgetful functor from modules to sets has a left adjoint, so the final module must have one element and the product in modules have the cartesian product as its underlying set. ut the initial module 0 does not have underlying set the empty set, so the forgetful functor can t have a right adjoint. Example. The forgetful functor from topological spaces to sets has both left and right adjoints. These are given by: the discrete and the indiscrete topology. (Convince yourself this is true.) Note the existence of these adjoints means that for any limit or colimit construction you want to perform in topological spaces, you already know what the underlying set must be. 2 Example. Consider a map of rings φ : R S. Then given any S-module N, we can regard it as an R-module by the formula r n := φ(r) n. It s common to write the resulting R-module as N R ; this is a forgetful functor from S-modules to R-modules. For an R-module M, we have a natural isomorphism Hom S (M R S, N) = Hom R (M, N R ) Thus R S gives a left adjoint of the forgetful functor. Actually, there s also a right adjoint: Hom R (N R, M) = Hom S (N, Hom R (S, M)) 2 I find it curious that the existence of the discrete and indiscrete topologies tell you anything at all.

5 COMMUTATIVE ALGERA LECTURE 2: MORE CATEGORY THEORY 5 It follows from the existence of these adjoints that taking limits or colimits commutes with forgetting the S-module structure. Note in particular that, taking R = Z, the underlying abelian group of a limit or colimit is the limit or colimit of underlying abelian groups. Example. Consider an R-module, M. Then for any other R-modules, L, N, we have Hom R (L R M, N) = Hom R (L, Hom R (M, N)) Thus we learn that R M preserves colimits, and in particular right exact sequences, and that Hom R (M, ) preserves limits, and in particular, left exact sequences. Example. Using the above adjunction both ways, we deduce Hom R (M, Hom R (L, N)) = Hom R (L, Hom R (M, N)) This doesn t look like an adjunction. However, let s think more carefully about Hom R (, N). It is contravariant, which means there s two ways of thinking of it: Thus we can rewrite the above equality as Hom R (, N) : R mod op R mod Hom R mod op(n, ) : R mod R mod op Hom R mod op(hom R mod op(n, L), M) = Hom R (L, Hom R (M, N)) Now you see that again Hom R (, N) is a left adjoint. Thus it takes left-exact sequences in R mod op aka right exact sequences in R mod to left-exact sequences in R mod. 3 You might call this an op-fuscation of the previous example. Finally, let s prove lemma 2. We ll do the right adjoints. Say we have some diagram of objects d α in D; assume lim d α exists. We compute: Hom C (c, g(lim d α )) = Hom D (f(c), lim d α ) = lim Hom D (f(c), d α ) = lim Hom C (c, g(d α )) The first and third equality use the adjunction; the second holds by definition of limit. Comparing the left and right most terms, we learn (by the definition of limit) g(lim d α ) = lim g(d α ). 3 We haven t really explained why exact sequences in R mod op is a thing that makes sense. Here s why: taking opposite category interchanges limits and colimits, so it s clear that R mod op has a zero (= initial and final) object, and pushouts are interchanged with pullbacks. So we can ask for pushout-pullback squares with a zero in a corner.