This is coming to you from Yiwei LI (PhD, Applied math), Taiyuan University of Science and Technology  (TYUST) Taiyuan, China

It's going on here for the third round of learning of Birkar's BAB-paper (v2), with scenarios of chess stories. No profession implications. 

This part appears over omitted.

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Th 2.15    Th 1.8            ♖    ♘

     ↓   ↖     ↓

Th 1.1      Th 1.6            ♔    ♗

    Mathematics vs Palace stories.(v2)

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Note: technical theorem is not on the board.

♙ ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁺⁻⁰ 1

(continued) camera level one = {R, S, '} ==> camera of level two = {Rs, Rs', R', ...}.

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Step 7, Para three ——

Assume now that (X, B⁺) is not lc over z = f(S).

---- This is to setup the assumption for a proof by contradiction.

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By the previous paragraph, (X, B⁺) is lc near S and S is a non-klt centre of this pair.

---- This is a nice summary, so that one can judge what is "lc near S".

---- Basically, for certain pair, Say (X, B), one can define Ks + Bs by (Kx + B)|s.

---- Or, given Ks + Bs, one can recover the original pair by "inversion of adjunction", to show Ks + Bs = (Kx + B)|s.

---- The divisor S is required to be a "non-klt centre" of the original pair.

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New comment: These old comments appear strange to me.

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On the other hand, (X, Γ) is plt with ⌊ Γ ⌋  = S, so if u > 0 is sufficiently small, then (X, (1 - u)B⁺ + uΓ) is plt near S and S is a non-klt centre of this pair and no other non-klt centre intersects S.

---- In combination of last sentence, if (X, B + ) is lc near S and (X, Γ) is plt near S, then the pair of (small) convex combination is still plt near S.

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Comment (wrong intuition): At the first glance, one might feel ⌊(1 - u)B⁺⌋ = 0 and ⌊uΓ⌋ = 0, for their coefficients are smaller than 1 ——

---- This is really the case. (No kidding).

---- But, there is a re-organization matter, however, for the sum of the two terms, before taking the floor operation.

---- That is, (1 - u)S + uS = S, while S has the coefficient 1.

---- For this aspect, u can be any number in [0, 1].

---- The small positive u is required by some other aspects.(to be hunted).

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New comment:  ⌊ (1 - u)B⁺ + uΓ ⌋ = S ?

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Then since (X, B⁺) is not lc over z, the non-klt locus of (X, (1 - u)B⁺ + uΓ) has at least two connected components (one of which is S) near the fibre f⁻1 {z}.

---- Translation: if the original pair is not lc over z, then the (plt convex) combined pair has its non-klt locus possessed at least two connected components near f⁻1 {z}.

---- To illustrate this in a simple way, take the boundary as the pair ——

    B⁺      ~    u(B⁺ )Γ

not | lc          not | simple

    z        ~      f⁻1 {z}

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Note:  u(B⁺ )Γ, understood in context, is the homemade notation for (X, (1 - u)B⁺ + uΓ).

Note: I call a (plt) pair is not simple, if its non-klt locus has least two connected components near f⁻1 {z}.

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New comment: The arise of (X, (1 - u)B⁺ + uΓ) is not apparent.

This contradicts the connectedness principle [25, Theorem 17.4] as - (Kx + (1 - u)B⁺ + uΓ) = - (1 - u) (Kx + B⁺ ) - u(Kx + Γ) ~ R -u(Kx + Γ) ~ R uαM - u(Kx + Γ) /Z is ample over Z.

---- That is, the defence form of u(B⁺)Γ is ample over Z, which expects u(B⁺)Γ simple near f⁻1 {z} by the connectedness principle.

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New comment: It is not apparent why - (1 - u) (Kx + B⁺ ) disappears and why uαM appears.

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Therefore, (X, B + ) is lc over z.

---- This is the close statement.

---- It's needed.

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New summary: This part appears over omitted.

↑↓ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ⌈ ⌉ ∨∧∞Φ⁺⁻⁰ 1

Calling graph for the technical theorem (Th1.9) ——

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Th1.9

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[5, 2.13(7)]   Lem 2.26   Pro4.1   Lem2.7

                                                          |

.....................................................Lem2.3   

Note: Th1.9 is only called by Pro.5.11, one of the two devices for Th1.8, the executing theorem.

Pro4.1                                                    

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[5, ?]   [37, Pro3.8]   [5, Lem3.3]   Th2.13[5, Th1.7]   [16, Pro2.1.2]  [20]  [25, Th17.4]

Completed notes of the first round learning for v2 Pro.4.1 are packaged on RG.

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Special note: Original synthesized scenarios in Chinese for the whole proof of v1 Th1.7, the technical theorem.

*It's now largely revised* due to new understandings.

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See also: Earlier comments in Chinese* (v1).

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It is my hope that this action would not be viewed from the usual perspective that many adults tend to hold.