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prose: fix diagrams/prose for lifts
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src/Cat/Morphism/Class.lagda.md

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@@ -26,15 +26,15 @@ arguments, so passing around functions like `∀ {x y} → Hom x y → Ω`
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as first-class objects can often lead to unsolved metavariable errors.
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To work around this, we define a class of morphisms in a category $\cC$
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as a single-field record type. This avoids the implicit instantiating
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issue from before, at the cost of a bit of code bloat.
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as a record type. This avoids the implicit instantiation issue from before,
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at the cost of a bit of code bloat.
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```agda
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record Arrows {o ℓ} (C : Precategory o ℓ) (κ : Level) : Type (o ⊔ ℓ ⊔ lsuc κ) where
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no-eta-equality
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field
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arrows : ∀ {x y} → Precategory.Hom C x y → Type κ
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is-tr : ∀ {x y} {f : Precategory.Hom C x y} → is-prop (arrows f)
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is-tr : ∀ {x y} {f : Precategory.Hom C x y} → is-prop (arrows f)
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open Arrows public
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```

src/Cat/Morphism/Lifts.lagda.md

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@@ -39,10 +39,10 @@ square like so:
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A && X \\
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\\
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B && {Y\text{.}}
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\arrow["f", from=1-1, to=1-3]
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\arrow["g", from=3-1, to=3-3]
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\arrow["u"', from=1-1, to=3-1]
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\arrow["v", from=1-3, to=3-3]
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\arrow["u", from=1-1, to=1-3]
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\arrow["v", from=3-1, to=3-3]
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\arrow["f"', from=1-1, to=3-1]
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\arrow["g", from=1-3, to=3-3]
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\end{tikzcd}\]
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~~~
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@@ -66,17 +66,17 @@ and $g$ *right lifts against* $f$ if for every commutative square
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A && X \\
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\\
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B && {Y\text{.}}
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\arrow["f", from=1-1, to=1-3]
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\arrow["g", from=3-1, to=3-3]
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\arrow["u"', from=1-1, to=3-1]
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\arrow["v", from=1-3, to=3-3]
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\arrow["u", from=1-1, to=1-3]
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\arrow["v", from=3-1, to=3-3]
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\arrow["f"', from=1-1, to=3-1]
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\arrow["g", from=1-3, to=3-3]
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\end{tikzcd}\]
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~~~
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there [[merely]] exists a lifting $w$.
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We can also talk about objects with left or right lifting properties.
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An object $P : \cC$ left lefts against a morphism $f$ if for every
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An object $P : \cC$ left lifts against a morphism $f$ if for every
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cospan $P \xto{u} X \xot{f} Y$, there merely exists a map $w : \cC(P,
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X)$ with $f \circ w = u$.
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@@ -343,15 +343,15 @@ $l$ and $k$ are both lifts of the outer square
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~~~{.quiver}
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\begin{tikzcd}
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a && b \\
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A && X \\
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\\
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c && d.
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\arrow["f", from=1-1, to=1-3]
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\arrow["u"', from=1-1, to=3-1]
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\arrow["l"', shift right, from=1-3, to=3-1]
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\arrow["k", shift left, from=1-3, to=3-1]
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\arrow["v", from=1-3, to=3-3]
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\arrow["g"', from=3-1, to=3-3]
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B && Y
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\arrow["u", from=1-1, to=1-3]
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\arrow["f"', from=1-1, to=3-1]
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\arrow["g", from=1-3, to=3-3]
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\arrow["k"', shift right=2, from=3-1, to=1-3]
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\arrow["l", shift left=2, from=3-1, to=1-3]
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\arrow["v"', from=3-1, to=3-3]
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\end{tikzcd}
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~~~
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