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--- blog:2023-03-04 [2023/03/04 20:22] – pzhou
+++ blog:2023-03-04 [2023/06/25 15:53] (current) – external edit 127.0.0.1
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 ==== L3 ====
+$\gdef\tto{\rightrightarrows}$
 Barr-Beck.
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+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.
+<del>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.</del>
+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.