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Stream: learning: questions

Topic: can a unique arrow makes more than one commuting diagram?

André Muricy Santos (Apr 24 2023 at 15:01):

very basic question. Suppose I have the below situation:
Screenshot-2023-04-24-at-16.49.00.png
The universal property of the product says that given any pair of arrows f1 and g1, there is a unique h such that the diagram commutes. But it says nothing else about h: in fact, why could there not be more pairs of arrows (in the picture f2 and g2) for which there is a unique arrow h' such that the diagram commutes AND h' happens to be h?

André Muricy Santos (Apr 24 2023 at 15:03):

I seem to remember that these uniquely determined arrows are sort of bound to the things that uniquely determine them, but only if they're "the only thing left to pick". So that if we had that g2 = g1, then we'd know that f2 = f1 also, like below:

Screenshot-2023-04-24-at-17.00.26.png

but then doesn't this trivialize currying? That is, if you can construct the operation of currying at all (something that "typechecks"), you automatically get a cartesian closed category because f simply has no choice but to be equal to curry g? This last part is very confused and I'm sorry about that, just not thinking productively anymore for today.

Morgan Rogers (he/him) (Apr 24 2023 at 15:24):

$f_1 = \pi_1 \circ h=f_2$ and $g_1=\pi_2 \circ h=g_2$.

Ralph Sarkis (Apr 24 2023 at 15:26):

Your first commutative diagrams says that $f_1 = \pi_1 \circ h = f_2$ and similarly for the $g$'s. One way to say this is that the universal property of the product is equivalent to there being a bijection $\mathrm{Hom}_{\mathbf{C}}(T,A\times B) \to \mathrm{Hom}_{\mathbf{C}}(T,A) \times \mathrm{Hom}_{\mathbf{C}}(T, B)$ for every $T$ (the second $\times$ is the Cartesian product of sets).

Morgan Rogers (he/him) (Apr 24 2023 at 15:26):

("The diagram commutes" means that the left hand and right hand triangles commute)

Ralph Sarkis (Apr 24 2023 at 15:27):

There was a similar discussion here.

André Muricy Santos (Apr 24 2023 at 15:46):

thank you both. It's obvious now. of course $h$ uniquely determines $f$ and $g$, they are equal to the composites of $h$ with the projections.