General #
First Part #
If γ is a path, then the image of [i/(d+1), (i+1)/(d+1)] under γ split at (d+1)/(n+1) is
the
image of [i/(n+1), (i+1)/(n+1)] under γ
If γ is a path, then the image of [i/(d+1), (i+1)/(d+1)] under γ split at (d+1)/(n+1) is
the
image of [i/(n+1), (i+1)/(n+1)] under γ.
Second Part #
When γ is a dipath, an we split it on the intervals [0, 1/(n+1)] and [1/(n+1), 1], then the image of γ of [(i+1)/(n+1), (i+2)/(n+1)] is equal to the image the second part of γ of [i/n, (i+1)/n]
When γ is a dipath, an we split it on the intervals [0, 1/(n+1)] and [1/(n+1), 1], then the image of γ of [(i+1)/(n+1), (i+2)/(n+1)] is equal to the image the second part of γ of [i/n, (i+1)/n]. Version with interval of real numbers
When γ is a dipath, an we split it on the intervals [0, (d+1)/(n+1)] and [(d+1)/(n+1), 1], then the image of γ of [(i+d.succ)/(n+1), (i+d.succ+1)/(n+1)] is equal to the image the second part of γ of [(i/(n-d), (i+1)/(n-d)].
When γ is a dipath, an we split it on the intervals [0, (d+1)/(n+1)] and [(d+1)/(n+1), 1], then the image of γ of [(i+d.succ)/(n+1), (i+d.succ+1)/(n+1)] is equal to the image the second part of γ of [i/(n-d), (i+1)/(n-d)].
Mixed Parts #
Splitting a dipath γ at [0, k/n] and then at [0, 1/k] is the same as splitting it at [0, 1/n].
Splitting a dipath [0, (k+1)/(n+1)] and then [1/(k+1), 1] is the same as
splitting it [1/(n+1), 1] and then [0, k/n]
Splitting a dipath [(k+2)/(n+2), 1] is the same as splitting it [1/(n+2), 1] and then [(k+1)/(n+1), 1]
Trans Parts #
If γ₁ and γ₂ are two paths, then the first part of γ₁.trans γ₂ split at 1/2 is γ₁
If γ₁ and γ₂ are two paths, then the second part of γ₁.trans γ₂ split at 1/2 is γ₂
If γ₁ and γ₂ are two paths, then the first part of γ₁.trans γ₂ split at 1/(2n + 2) is
the
same as γ₁ split at 1/(n + 1).
If γ₁ and γ₂ are two paths, then
γ₁.trans γ₂ --> [1/(2n + 4), 1] --> [0, (2n + 2)/(2n + 3)] (so taking [1/(2n + 4), (2n + 3)/(2n + 4)])
is the same as
γ₁ --> [1/(n+2), 1], added to γ₂ --> [0, (n+1)/(n+2)]
If γ₁ and γ₂ are two paths, then
γ₁.trans γ₂ --> [1/(2n + 4), 1] --> [(2n+2)/(2n+3), 1] (so to [(2n+3)/(2n+4), 1])
is the same as
γ₂ --> [(n+1)/(n+2), 1]
If γ₁ and γ₂ are two paths, then γ₁.trans γ₂ evaluated at 1/(2n+2) is the same as
γ₁ evaluated at 1/(n+1).
If γ₁ and γ₂ are two paths, then γ₁.trans γ₂ --> [1/(2n+4), 1] evaluated at
(2n+2)/(2n+3) is the same as
γ₂ evaluated at (n+1)/(n+2).