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This is an exercise from Algebra of Programming which I'm reading for self-study. Below, $T$ is the initial algebra of $F$.

enter image description here

What I've tried:

I can construct $h \circ F (\pi_2) : F(A \times B) \rightarrow B$. Then we take the product $\langle g, h \circ F(\pi_2) \rangle : F(A \times B) \rightarrow A \times B$. We can then take the catamorphism $cata(\langle g, h \circ F(\pi_2) \rangle) : T \rightarrow A\times B$ and need to show that $f = \pi_1 \circ cata(\langle g, h \circ F(\pi_2) \rangle)$.

I've tried using the fact that the diagram above commutes to get the equality above, to no avail. Any suggestions?

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    $\begingroup$ What are $outl$ and the banana brackets around $k$? $\endgroup$ Commented Jul 8, 2022 at 2:01
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    $\begingroup$ Hint: prove the stronger statement $(f,cata(h))=cata(\langle g,h\circ F(\pi_2)\rangle),$ using the uniqueness in the definition of cata. $\endgroup$ Commented Jul 8, 2022 at 8:48
  • $\begingroup$ @ColinMcQuillan thanks! I was able to find a solution with your hint. $\endgroup$
    – abeln
    Commented Jul 8, 2022 at 14:11

1 Answer 1

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The solution as per @Colin McQuillan's hint above:

We show that the product $\langle f, cata(h) \rangle$ is equal to the catamorphism $cata(\langle g, h \circ F(outr) \rangle)$.

By the universal property of catamorphisms, it suffices to show that

$\langle f, cata(h) \rangle \circ \alpha = \langle g, h \circ F(outr) \rangle \circ F \langle f, cata(h) \rangle$

On the left, we have

$\langle f, cata(h) \rangle \circ \alpha = \langle f \circ \alpha, cata(h) \circ \alpha \rangle = \langle g \circ F \langle f, cata(h) , h \circ F (cata(h)) \rangle \rangle$

Notice that $F(cata(h)) = F outr \circ F \langle f, cata(h) \rangle$, so

$\langle g \circ F \langle f, cata(h) , h \circ F (cata(h)) \rangle \rangle = \langle g \circ F \langle f, cata(h) \rangle, h \circ F outr \circ F \langle f, cata(h) \rangle\rangle = \langle g, h \circ F outr \rangle \circ F \langle f, cata(h) \rangle$

as needed. $\blacksquare$

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