start hcomp post

plugin/remark-agda.ts

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     const htmlname = parse(path).base.replace(/\.lagda.md/, ".html");
 
+    console.log("Output file:", outputFile);
     const doc = readFileSync(outputFile);
 
     // This is the post-processed markdown with HTML code blocks replacing the Agda code blocks
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     });
 
     let idx = 0;
+    try {
       visit(tree, "code", (node) => {
         if (!(node.lang === null || node.lang === "agda")) return;
 
+        if (idx > collectedCodeBlocks.length) {
+          throw new Error("failed");
+        }
+
         node.type = "html";
         node.value = collectedCodeBlocks[idx].contents;
         idx += 1;
       });
+    } catch (e) {
+      console.log(
+        "Mismatch in number of args. Perhaps there was an empty block?",
+      );
+    }
   };
 };
 

src/content/posts/2024-09-18-hcomp/index.lagda.md

@@ -0,0 +1,87 @@
+---
+title: Examples of hcomp
+slug: 2024-09-18-hcomp
+date: 2024-09-18T04:07:13-05:00
+tags: [hott, cubical]
+draft: true
+---
+
+**hcomp** is a primitive operation in cubical type theory.
+
+```
+{-# OPTIONS --cubical --allow-unsolved-metas #-}
+module 2024-09-18-hcomp.index where
+open import Cubical.Foundations.Prelude hiding (isProp→isSet)
+open import Cubical.Core.Primitives
+```
+
+Intuitively, hcomp can be understood as the composition operation.
+
+```
+path-comp : {A : Type} {x y z : A} → x ≡ y → y ≡ z → x ≡ z
+path-comp {x = x} p q i =
+  let u = λ j → λ where
+    (i = i0) → x
+    (i = i1) → q j
+  in hcomp u (p i)
+```
+
+## Example: $\mathsf{isProp}(A) \rightarrow \mathsf{isSet}(A)$
+
+Suppose we want to prove that all mere propositions (h-level 1) are sets (h-level 2).
+This result exists in the cubical standard library, but let's go over it here.
+
+```
+isProp→isSet : {A : Type} → isProp A → isSet A
+isProp→isSet {A} A-isProp = goal where
+  goal : (x y : A) → (p q : x ≡ y) → p ≡ q
+  goal x y p q j i = -- ...
+```
+
+Now let's construct an hcomp. In a set, we'd want paths $p$ and $q$ between the same points $x$ and $y$ to be equal.
+Suppose $p$ and $q$ operate over the same dimension, $i$.
+If we want to find a path between $p$ and $q$, we'll want another dimension.
+Let's call this $j$.
+So essentially, we want a square with these boundaries
+
+* the left is $\mathsf{refl}_x$
+* the right is $\mathsf{refl}_y$
+* the bottom is $p(i)$
+* the top is $q(i)$
+
+Our goal is to find out what completes this square.
+Well, one way to complete a square is to treat it as the top face of a cube and use $\mathsf{hcomp}$.
+
+Remember:
+
+* $i$ is the left-right dimension, the one that $p$ and $q$ work over
+* $j$ is the dimension of our final path between $p \equiv q$.
+  Note that this is the first argument, because our top-level ask was $p \equiv q$.
+* Let's introduce a dimension $k$ for doing our $\mathsf{hcomp}$
+
+We can map both $p(i)$ and $q(i)$ down to a square that has $x$ on all corners and $\mathsf{refl}_x$ on all sides.
+The method is this:
+
+* the bottom face $(k = \mathsf{i0})$ is the constant $x$
+* the left face $(i = \mathsf{i0})$ is _also_ the constant $x$
+* the right face $(i = \mathsf{i1})$ is trickier.
+  We have $x$ on the bottom 2 corners, but $y$ on the top two corners.
+  Fortunately, $\mathsf{isProp}(A)$ tells us that $x$ and $y$ are the same, so $x \equiv y$.
+  This means we can define this face as $\mathsf{isProp}(A, x, y, j)$.
+* the same logic applies to the front face $(j = \mathsf{i0})$ and back face $(j = \mathsf{i1})$.
+  We can use $\mathsf{isProp}$ to generate us some faces, except using $x$ and $p(i)$, or $x$ and $q(i)$ as the two endpoints.
+
+Now we can try to find the top face $(k = \mathsf{i1})$:
+
+```
+    let u = λ k → λ where
+      (i = i0) → A-isProp x x k
+      (i = i1) → A-isProp x y k
+      (j = i0) → A-isProp x (p i) k
+      (j = i1) → A-isProp x (q i) k
+    in hcomp u x
+```
+
+This type-checks! Let's move on to a more complicated example.
+
+## Example: $\Sigma \mathbb{2} \rightarrow S^1$