• learn what is Barr-Beck (don't worry about Lurie, I am not $\infty$ yet).

$\gdef\colim{\text{colim}}$

Let me follow Branter's note . And, Akhil Matthew's Serre Criterion for affiness.

Category warm-up.

• What is a left-adjoint functor? $\otimes$, $i^*$ (restriction to open set)
• What is a colimit? (taking cokernel of $A \to B$, taking coequalizer)
• Who preserves colimit? $Hom( colim_i A_i, B) = \lim_i Hom(A_i, B)$ by definition. Thus, if $L$ is a left-adjoint, we have $$ Hom(L (colim A_i), B) = Hom( (colim A_i), R B) = lim Hom(A_i, RB) = lim Hom(LA_i, B) = Hom(colim L A_i, B) $$
• That was too abstract. What is an example? No, remember it, like 'colimit commute with left-adjoint', say it 100 times.
• If $P$ is a compact projective, then $Hom(P, -)$ commute with all colimit? Why?
  • $P$ is compact, means that $Hom(P, -)$ preserves filtered colimit (can there be an example, where $P$ does not

What is a filtered colimit? It is a colimit over a filtered category, where every two objects can map to a third common object, and every two morphism (within the same hom space) can be composed with a third, to coequalize. As written, it is obvious that, pushout square, direct sum of stuff (even finitely many), cokernel (a special kind of pushout) are not filtered colimit.

see this note

what is a directed set? it is a set with partial ordering, such that every two elements has a common downstream guy. So, it is less general than a filtered set, where morphism between two objects can be not just $<$ relation.

$\Q$ is a direct limit (union) of copies of $\Z$. Indeed, it is like $\Q = \cup_{n > 0} \frac{1}{n} \Z $. As a $\Z$-module. very nice.

What is a localization of a ring? where the ring is viewed as a module over itself, and we allow for bigger and bigger denominators? maybe.

what does flat module mean? tensor is good?

I have this conflicting intuition: for a filtered diagram, every two nodes eventually will meet and will stay together ever after, but there will be more and more nodes as you 'go to the right, go to infinity'. how can that happen? well, it is just like life, each individual will die, but the human society lives on. fine.

why filtered colimit preserves flatness? If $M_i$ are flat modules, and $M = \colim_i M_i$ is a filtered colimit, we need to show that for any $A \to B$ injection, we have $M \otimes A \to M \otimes B $ is still injective. Why? suppose we have $\sum_i m_j a_j$ sent to $\sum_j m_j b_j$ which is zero. Here we have finitely many $m_j$ involved in the colimit, each of them represent a coherent string of showers of modules in the system. The point is that, all the $m_j$ have a common home somewhere downstream. We can use property of that common home.

Fact: all colimit preserves right-exatness. namely, if we have a surjection $M_i \to N_i$, then the direct sum of them still is a surjection. Why arbitrary colimit does not perserve left-exactness? Here is an example: take $0 \into \Z$, and $\Z \into \Z$, and $\Z \into \Z$, and consider the pushout diagram from the first to the 2nd,3rd. then the termwise pushout gives me $\Z^2 \to \Z$, but it doesn't preserved left exactness.

You see, colimit is a useful notion, without which we cannot talk about union, cokernel, many things. filtered colimit is another useful notion, with good properties, but too strict. They are different.

Let me do this exercises:

• If $P$ is compact, then $\Hom(P, -)$ commute with filtered colimit. yes, definition. And only filtred colimit. For example, consider $P = \Z/2\Z$ as $\Z$-mod, and consider the cokernel of $\Z \xto{2} \Z$, we get $\Z/2$ as the 'colimit of the pushout'. However, if we apply the $Hom(P, -)$, termwise, we get $0 \xto{2} 0$, hence the pushout is $0$. You can say, the failure is because we didn't use derived hom.
• If $P$ is projective, then, there is no higher hom. $Hom(P, -)$ is right exact.
• finite limit and filtered colimit commute in SET.

Warm-up: what is limit of a category in set? it is like taking global section of a sheaf, but much much more derived. it involves choosing for each object $i \in I$, a corresponding value $x_i \in X(i)$, such that they are compatible along maps. sometimes, it is not even possible, like, you have two arrows $a,b: i \to j$, but there is no $c_i \in X(i)$ such that $a(c_i) = b(c_i)$. bad.

Warm-up: what is the filtered colimit in Set? It is a disjoint union of the valued set, quotient out by some realtion.

Warm-up: what is an arbitrary colimit in Set? For example, how do you take the 'pushout' in SET? Well, the best thing that I can think of is the connecting dots using lines, and take connected components.

Warm-up: now that I know, in set, limit = global section; colimit = gluing, and filtered colimit = gluing without using lines to connect things. then if I have two indexing categories, a filtered cat $I$ and a finite cat $J$, and a bi-functor $D(i,j)$, then I can either take $J$ direction global section, then patch it up, in the I direction. or I can take the I-direction patching, then take the resulting global section. The second approach is apparently more complicated. The idea is that, for the finite limit of the filtered colimit, for each node $j$, we choose some guy in the termwise colimit, so that they are compatible in the colimit-sense relations. Now, we take representatives back in the finite place. they may each map to some different, but finitely many $i(j)$. We can first let them meet-up downstream at some common node $i$. but then, the arrows between them may not match-up, but we know they evnetually match-up, and there are only finitely many arrows, so we flow downstream, until they one by one all become green-check. So, then, we finished constructed a lift from the global section of the filtered colimit to something easy and nice.

Notation: right-cone of a category $I$: just add a terminal object to $I$.

Say $F: C \to D$ is a functor, we say $F$ preserves and reflect colimit, meaning: suppose we have a candidate colimit diagram in $C$, and take its image in $D$. Then, the diagram is colimit (right exact) in $C$, if and only if it is colimit in $D$. Note that, we don't assume whether all diagrams can be cocompleted to a colimit diagram, either in $C$ and in $D$. I am only saying, the amount of colimit diagram are the same in $C$ and in $D$.

Suppose $U: C \to Set$ is a faithful functor (no hom gets killed, hence no objects get killed), and suppose $U$ matches filtered colimit and finite limit on both sides (no add, no subtract), then finite limit and filtered colimit in $C$ commute. Application: $C$ is hte abelian group of $R$-module. Indeed, when I think of filtered colimit and limit of an R-module, I think of hte underlying set.

Finite colimit of compact object is compact. Because say $(B_j)$ is a filtered colimit over $J$, and $A_i$ is a finite colimit over $I$, and we have $A=colim A_i$, $B = colim B_j$, we want to know $$ Hom(A, colim_j B_j) = lim_i Hom(A_i, colim_j B_j) = lim_i colim_j Hom(A_i,B_j) = colim_j lim_i Hom(A_i, B_j) = colim_j Hom(A, B_j)$$ in the middle, I used the fact that $Hom(A_i, B_j)$ is a set.

A module over a ring $R$ is compact, if and only if it is finitely presented. I can understand finite generation, but don't understand finite presentation. Let me show that such thing are compact. First, $Hom_R(R, -)$ is the same as forgetful functor to abelian group. it is the right-adjoint to Free. $R$ is projective. I want to say it commutes with all colimit. finite colimit of compact is compact, hence finite presented is compact. Finally, why every compact is of this form? Can we say, finitely presented sub-objects? Suppose we fix the set of generators, and we increase the amount of relations. The thing with less relations maps to the thing with more relations. So, we can look at the sub-category of finitely presented modules, and look for morphisims that map to this given compact guy. all possible such morphisms. these guys forms a filtered colimit. We know, compact guy is a filtered colimit of the fin-pres guys, so we can take the identity morphism. So, this morphism shows up at certain point of the $id_C \in filt-colim(C, C_i)$, so we can obtain $C_i$ as a finite exact truncation of $C$.

what is a reflexive pair? think of two deformation retraction of $A \rightrightarrows B$, where $B \into A$. ok, it is what it is.

What does co-equalizer mean? Can we say $B$ is the co-equalizer, since everything maps to $B$? well, but how does $A$ map to $B$, we had two candidate. We need to quotient a bit. again, let's use lines to link the points, then we take the connected component. check 1, it is canonical. yes, this reflexive pair does not add any new arrow. it just witness the fact that, each point in B is hit by at least one red and one green lines, and there exists a choice of blue lines in reverse direction that revert both red and green.

$\gdef\tto{\rightrightarrows}$ Barr-Beck.

Monad. A monad $T$ in $C$, is an endofunctor with some extra data satisfying some conditions. monad needs a unit and a composition. An endfunctor tells you how to move objects and morphism in a category, so that composition and stuff doesn't break. A natural transformation $\eta$ between $1$ and a endo $T$, gives a bridge( could be 0), such that for each object $\eta_X: X \to T(X)$ is a bridge, and we have many many commutative squares.

Usually, you give me an adjunction pair $L: C <-> D: R$, we get a monad $T:= RL$ acting on $C$, and we can ask, can we reconstruct $D$ using $R(D)$ as a $Mod_T(C)$? We view $C$ as some weak guy, and we view having these extra data (monad action) constraint what objects you can use.

Sometimes, the dream fails. For examples $L = free: Set <-> Top: R = forget$. Then $R$ can fail to be conservative, i.e. $R$ does not match isomorphisms, it can sends non-isomorphic bijection to bijection which is isom on Set.

What's the crude Barr-Beck? It says, if you have an adjuction like $$ L: C \leftrightarrow D: R$$ and constructed the monad $T = RL$, then if two conditions are satisfied, then the comparison morphisms $R: D \to Mod_T(C)$ is an equivalence. One condition is really easy to check: $R$ reflect isomorphism; the other condition is harder, D admits reflexive co-equalizer (like $D$ admits colimit), and $R$ preserves reflexive coequalier (like$R$ preserves cokernel)

Here is the weirdest definition in the world: split coequalizer. I bearly understand reflexive pair, and its coequalizer; now I need to do a new one. We see a parallel morphism $X_1 \rightrightarrows X_0$ split, if there exists coequalizer $h: X_0 \to X_{-1}$, and arrows $s: X_{-1} \to X_0, t: X_0 \to X_1$. Such that

• $h$ coequalizes $d_0, d_1$ (but may not be the co-equalizer), $s$ is the embedding (section of $h$). rather we can view $X_{-1}$ as a subset of $X_0$, and $s=id$ and $h$ is the projection.
• $t: X_0 \to X_1$ can be thought of as an embedding, and $d_0: X_1 \to X_0$ is a projection.
• So, this part is easy to understand: that as vector space, we have nested subspaces with well-chosen complements $$ X_1 = V_1 \oplus V_0 \oplus V_{-1}, \quad X_0 = V_0 \oplus V_{-1}, \quad X_{-1} = V_{-1}. $$
• What's the condition on the remaining arrow? $d_0$ just kills $V_1$ and is identity on the rest. $d_1: X_1 \to X_0$ (viewed as map to itself), such that when restricted on $X_0$ projects to $X_{-1}$.

I don't know how to generalize this to many steps.

Let's prove the Barr-Beck (crude version). We have a pair of adjoint functors $L: C <-> D;R$, and $T =RL$, and we assume $R$ is conservative, and $D$ admits reflexive coequalizer, and $R$ preserve those.

Let's assume we have additive category. no need to assume abelian category. If $D$ is an abelian category, and $R$ preserve small colimit, then these conditions are automatically true.

We want to know if $\wt R: D \to Mod_T(C)$ by $c \mapsto R(c )$ is an equivalence or not.

• Step 1: we need to construct a fancy left-adjoint $\wt L: Mod_T(C) \to D$. No, this is not the composition of forget to $C$ and then apply $L$, because hom in $Mod_T(C)$ and in $C$ are quite different. We anticipate $\wt L$ is the inverse, so it should behave more like $R^{-1}$. For example, we expect it to do $\wt L\wt R(d) = d$. However, not every T-module in $C$ is written as $\wt R(d)$. But, no worries, we should be able to resolve, and then strip off the $R$ from the resolution. Suppose $A \in Mod_T(C)$, we have resolution $$ T(T(A)) \rightrightarrows T(A) \xto{m} A $$ So, we should get

$$ LRL(A) \rightrightarrows L(A) \xto{m} \wt L(A) $$ In other words, we all know that going around the adjoint loop, picks up too much junk, we need to clean up. Hold on. This functor as stands is just some functor, how do we know that it is THE adjoint? And, even worse, why do we have coequalizer? We say, this pair is a reflexive pair, since $LA \to LRLA$ we can insert $1\to RL$ in the middle. Need to check if $$ LA \xto{L(1 \to RL)A} LRLA \xto{(LR \to 1)LA} LA$$ is identity. who says so? counit axioms of adjunction says so. The other thing is $A \to TA \to A$ unit compose with action is identity, so again ok. We indeed have a reflexive pair. So, using the requirements, we know that this reflexive guy admits a co-equalizer, and $G$ preserves it.

There is a related fact about monad $T$. We have $$ TA \xto{(1 \to T)(TA)} TTA \xto{(TT \to T)A} TA$$ this has nothing to do with $A$, purely axioms of monad $T \xto{(1\to T)T} TT \xto{m} T$ is identity. Good, we also have $$ TA \xto{T(1 \to T) TTA} \xto{T(TA \to A)} TA$$ which has nothing to do with the left most $T$, which is again identity.

And further more $(1 \to T)T = T(1 \to T)$, so the two sections are actually the same. So, we indeed have a common section. Thus $TTA \rightrightarrows TA$ is a reflexive pair. WOW, this is wrong, the two maps $T \to TT$ are different, it matter where you insert this $T$. So, we don't have a reflexive pair!

Instead, we have a split pair. $$ TTA \tto TA \to A $$ How? $A \to TA$ exists using $1 \to T$. And $TA \to TTA$ append on the leftmost $1 \to T$. So we have the desired embedding. And, it is easy to check that, appending $1$ on the left and actual merge $TA \to A$ commute. Hence this is a split pair.

Hold on, again, this is split in $C$, not in $Mod_T(C)$, indeed, the splitting morphisms are not $T$-module. The action of $T$ on bare $A$ is using real data, and the action of $T$ on $1\cdot A$ is formal.

We need to check that, this sequence, is indeed a coequalizer in $Mod_T(C)$. Not hard.

We can do the sharper version. We know that this pair, $LRLA \tto LA$ in $D$, when apply $R$, becomes $TTA \to TA$ in $C$. it is a split coequalizer in $C$, so the original pair in $D$ admits a coequalizer $\wt L A$ in $D$, such that $R \wt L A \cong A$ (since $R$ matches such coequalizer).

• Step 2: show that $id_{Mod_T(C)} \to \wt R \wt L$ is an equivalence.
• Step 3: show that $\wt L \wt R \to id_D$ is an equivalence.

So, what did I learn? nothing, just write down the correction adjoint with correction.