Proof that ⋁ is not Right adjoint to Diagonal Functor on 2

category-theory

I was reading the book Diagrammatic Immanence: Category Theory and Philosophy which has one chapter on Category Theory following each philosophical chapter: one on Spinoza, another on Peirce and finally one on Deleuze. In the last chapter On Adjunctions and Topoi, the author Rocco Gangle gives the example of the Right Adjunct to the diagonal functor from the unique arrow category 2. The right adjoint is logical conjunction if the domain of the unique arrow is understood as Falsity and the range is Truth. Here is my version of the diagram.

Right Adjoint to Delta

So I thought as a quick exercise I'd try to show that Or could not be the right adjoint to the same diagonal functor. This seemed nice and simple as one could easily just list all the cases.
Probably because I am a bit rusty, this has been taking me quite a lot longer than I was anticipating!

Here's the diagram of the functor for Or (⋁) which should be the left adjunction to the diagonal functor.

Left Adjoint to Delta

So for the (counterfactual) case ⋁ were to be a right adjoint we would require

  1. there to be an isomorphism of Hom-Sets
    $H_{2\times2}(\Delta(A),(c,d)) \cong H_2(A, c \lor d)$
  2. that is natural in $\Delta$ and $\lor$
    (I am not quite sure what that means)

So I thought I'd just work out the homsets. (Beware if you are in night-time mode: ⚫︎ is a full circle corresponding to the black circle in the picture, the domain of the h arrow or False).

$$
H_{2\times2}(\Delta⚫︎, ○○) = \{\langle h,h \rangle \} \cong H_2(⚫︎, ○ \lor ○) = \{ h \} \\
H_{2\times2}(\Delta⚫︎, ⚫︎⚫︎) = \{\langle 1_⚫︎,1_⚫︎\rangle \} \cong H_2(⚫︎, ○ \lor ○) = \{ h \} \\
H_{2\times2}(\Delta⚫︎, ○⚫︎) = \{\langle h,1_⚫︎ \rangle \} \cong H_2(⚫︎, ○ \lor ⚫︎) = \{ h \}\\
H_{2\times2}(\Delta⚫︎, ⚫︎○) = \{\langle 1_⚫︎,h \rangle \} \cong H_2(⚫︎, ⚫︎ \lor ○) = \{ h \}\\
H_{2\times2}(\Delta○, ○○) = \{\langle 1_○,1_○ \rangle \} \cong H_2(○, ○ \lor ○) = \{ 1_○ \}
$$

But those seem to work out. So where does it break down?

Best Answer

To speak in the 1960ies existentialist way: I was blinded by nothingness.

Or in more mundane speak: I ignored all relations with empty hom-sets, especially these two:

$$ H_{2\times2}(\Delta○, ○⚫︎) = \{ \} \ncong H_2(○, ○ \lor ⚫︎) = \{ 1_○ \} \\ H_{2\times2}(\Delta○, ⚫︎○) = \{ \} \ncong H_2(○, ⚫︎ \lor ○) = \{ 1_○ \} \\ $$

Thanks to S.C for pointing that out.

There is a third one but that is not a counterexample

$$ H_{2\times2}(\Delta○, ⚫︎⚫︎) = \{ \} \cong H_2(○, ⚫︎ \lor ⚫︎) = \{ \} $$

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