Re: [Haskell-cafe] A quick question about distribution of finite maps.

Damn, you're right, my mistake. So, in short, there doesn't seem to be a good representation for (X^Y). But even if it was, we can be certain that #(X^Y) = (#X + 1) * (#X + 1)^(#Y) * (#X + 1)^(#Y) - 1 = (#X + 1)^(2 * #Y + 1) - 1 That makes it possible to calculate both #(Z -> Maybe X^Y) and #(OneOrBoth Z Y -> Maybe X): #(Z -> Maybe X^Y) = #(Maybe X^Y)^#Z = (#X + 1)^(2 * #Y * #Z + #Z) #(OneOrBoth Z Y -> Maybe X) = #(Maybe X)^#(OneOrBoth Z Y) = (#X + 1)^(#Y * #Z + #Y + #Z) and we still see that they are different, unless X is empty or Z ~ (). Thanks!

On 10 Jan 2021, at 22:10, Olaf Klinke wrote:

On Sat, 2021-01-09 at 23:45 +0100, MigMit wrote: But it won't be cartesian closed. If it were, then for any finite

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X and Y we should have

Maybe (X^Y) ~ () -> Maybe (X^Y) ~ OneOrBoth () Y -> Maybe X ~ (() -> Maybe X, Y -> Maybe X, ((), Y) -> Maybe X) ~ (Maybe X, Y -> Maybe X, Y -> Maybe X)

so

X^Y ~ (X, Y -> Maybe X, Y -> Maybe X)

No, my arrows and isomorphisms are all in the original category, not the Kleisli one — although the "X^Y" is the exponent in the Kleisli category.

I don't follow your argument. I still must be misinterpreting something.

Maybe (X^Y) ~ () -> Maybe (X^Y) -- because () is terminal in Hask ~ OneOrBoth () Y -> Maybe X -- OneOrBoth is the product in Kleisli Maybe ~ (Maybe X, Y -> Maybe X, Y -> Maybe X) -- universal property of coproduct

But how did you get to the next step:

X^Y ~ (X, Y -> Maybe X, Y -> Maybe X)

I think this can not hold for cardinality reasons:

1+X*((1+X)^Y)^2 /= (1+X)*((1+X)^Y)^2

Olaf