[數撚圍爐區] abc conjecture is still a conjecture (3)
吾不會數學
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每個符號都係一個function, 如果唔unique的咪唔係welk defined, 所以唔可以比符號佢
Suppose x ∈ S, α ∈ Cfx(S), and ϵ > 0. Let {αt
: |t| < ϵ}
be a holomorphic arc on Cfx(S) such that α0 = α. Then, shrinking ϵ if
necessary there exists a neighborhood U of x on X and a holomorphic
map A : U × ∆(ϵ) → Ce (X) such that (a) A(x, t) = αt for t ∈ ∆(ϵ),
(b) A(z, t) ∈ Cfz(X) whenever z ∈ U and (c) A(z
′
, t) ∈ Cfz
′(S) whenever
z
′ ∈ U ∩ S. Hence, for t ∈ ∆(ϵ), αet(z) := A(z, t) defines a holomorphic
section of πe : Ce (X) → X over U such that αet
|U∩S is a section of
ϖe : Ce(S) → S over U ∩ S. We have αet = αe + tξe + O(t
2
), where
αe(z) := A(z, 0), ξe(z) ∈ Pαe(z)
for z ∈ U, and ξe(z
′
) ∈ Pαe(z
′) ∩ Tz
′(S) for
z
′ ∈ U ∩ S.
: |t| < ϵ}
be a holomorphic arc on Cfx(S) such that α0 = α. Then, shrinking ϵ if
necessary there exists a neighborhood U of x on X and a holomorphic
map A : U × ∆(ϵ) → Ce (X) such that (a) A(x, t) = αt for t ∈ ∆(ϵ),
(b) A(z, t) ∈ Cfz(X) whenever z ∈ U and (c) A(z
′
, t) ∈ Cfz
′(S) whenever
z
′ ∈ U ∩ S. Hence, for t ∈ ∆(ϵ), αet(z) := A(z, t) defines a holomorphic
section of πe : Ce (X) → X over U such that αet
|U∩S is a section of
ϖe : Ce(S) → S over U ∩ S. We have αet = αe + tξe + O(t
2
), where
αe(z) := A(z, 0), ξe(z) ∈ Pαe(z)
for z ∈ U, and ξe(z
′
) ∈ Pαe(z
′) ∩ Tz
′(S) for
z
′ ∈ U ∩ S.
既然complex logarithm都可以multivalued
整個符號係equivalence class of all basis咪得
整個符號係equivalence class of all basis咪得
嚴格上黎講你要揀左branch先
如果你堅持唔揀branch 個output會係一個set 
不過我記得有一種方法處理呢類function但我唔記得點處理
你哋覺得topology 同number theory 對ug non pure math student 係咪必須