# What am I working with? [Inferring a theory in Category Theory using associativity of Cartesian Product]

In the category of sets, there is a "natural isomorphism," given three sets $A$, $B$, and $C$, from the set $(A \times B) \times C$ to the set $A \times (B \times C)$, where $\times$ is a Cartesian product. This observation suggests ta theorem in category theory. State and prove it. Similarly for disjoint union.

I think I'm a bit wet behind the ears here—but I'm not sure what kind of stuff I'm working with. That is, what would be parallel to $(A \times B) \times C$?

Let $A$, $B$, and $C$ be objects with direct products. We denote the objects in their respective direct products as $AB$, $AC$, $ABC$, etc.

• There exists an isomorphism between $(AB)C$ and $A(BC)$.
• $\hom{(AB,A)} \times \hom{(AB, B)} \times \hom{(ABC, C)} \cong \hom{(BC, B)} \times \hom{(BC, C)} \times \hom{(ABC, A)}$
• Consider sets whose elements are categorical objects. Then $(A \times B) \times C \cong A \times (B \times C)$.

I have a bunch floating around, and also suspect that these are all equivalent. The last two are trivial given the problem statement, right? I want to prove the first by showing (AB)C and A(BC) are both isomorphic to ABC.

I'm not really sure what theorem I'm supposed to be getting. Am I working with ismorphisms of sets of morphisms? Or isomorphisms of objects? In the category of sets, they're equivalent, I think.

Could someone point me in the right direction? Am I already in the right direction?

• I don't know enough to properly answer your question, but, this much is easy $((a,b),c)$ corresponds to $(a,(b,c))$ and this gives us the bijective correspondence of $(A \times B) \times C$ and $A \times (B \times C)$. That said, I'm not sure how you define $A \times B \times C$. I usually think of it as both $(A \times B) \times C$ and $A \times (B \times C)$. Anyway, I'm sure someone will help with this interesting question. Commented Jul 24, 2015 at 13:25
• Products are special objects in categories. They are defined by a pair of morphisms and a universal property. The cartesian product is the product in the category of sets. So how would you restate this theorem for a general category? Commented Jul 24, 2015 at 13:50
• @JohnDouma That's what the first bullet point is, right? That is, if $(A \times B) \times C \cong A \times (B \times C)$ in the category of sets, generally it would translate to something like: There exists an isomorphism between $(AB)C$ and $A(BC)$ where $XY$ denotes the direct product between objects $X$ and $Y$. Commented Jul 24, 2015 at 16:47

Products in Set are Cartesian products. Coproducts in Set are disjoint unions.

What is said about the category of sets concerning Cartesian products and disjoint unions must be stated and proved by you in general for categories that have binary products and binary coproducts respectively. Then concerning products and coproducts. It is not a matter of "using the associativity of the Cartesian product" as you suggest in the title of your question. It is a matter of stating and proving the more general analogue.

Let $\mathcal C$ denote a category that has binary products.

Multiplication can be interpreted as a bifunctor $F:\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$ prescribed by $\langle A,B\rangle\mapsto A\times B$ on objects and $\langle f,g\rangle\mapsto f\times g$ on morphisms.

Identifying the categories $\mathcal C\times (\mathcal C\times\mathcal C)$ and $(\mathcal C\times \mathcal C)\times\mathcal C$ both with $\mathcal C\times \mathcal C\times\mathcal C$ we have the functors:

• $F\circ\left(1\times F\right):\mathcal{C}\times\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$ prescribed by $\langle A,B,C\rangle\mapsto A\times\left(B\times C\right)$ on objects and likewise on morphisms.
• $F\circ\left(F\times1\right):\mathcal{C}\times\mathcal{C}\times\mathcal{C}\rightarrow\mathcal{C}$ prescribed by $\langle A,B,C\rangle\mapsto\left(A\times B\right)\times C$ on objects and likewise on morphisms.

To be shown is the existence of a natural isomorphism: $$\eta:F\circ\left(1\times F\right)\stackrel{\bullet}{\rightarrow}F\circ\left(F\times1\right)$$

This for products.

For coproducts the same with $\times$ replaced by $\sqcup$.

edit:

Let $\pi_{A}:A\times B\times C\rightarrow A$, $\pi_{B}:A\times B\times C\rightarrow A$ and $\pi_{C}:A\times B\times C\rightarrow A$ denote the projections.

Let $\rho_{AB}:\left(A\times B\right)\times C\rightarrow A\times B$ and $\rho_{C}:\left(A\times B\right)\times C\rightarrow C$ denote the projections.

Let $\tau_{A}:A\times B\rightarrow A$ and $\tau_{B}:A\times B\rightarrow B$ denote the projections.

Then we have the morphisms:

$\tau_{A}\circ\rho_{AB}:\left(A\times B\right)\times C\rightarrow A$, $\tau_{B}\circ\rho_{AB}:\left(A\times B\right)\times C\rightarrow B$ and $\rho_{C}:\left(A\times B\right)\times C\rightarrow C$

implying the existence of a unique $h:\left(A\times B\right)\times C\rightarrow A\times B\times C$ that satisfies: $\pi_{A}\circ h=\tau_{A}\circ\rho_{AB}$, $\pi_{B}\circ h=\tau_{B}\circ\rho_{AB}$ and $\pi_{B}\circ h=\rho_{C}$.

This $h$ can be shown to be invertible.

Also we can find $g:A\times\left(B\times C\right)\rightarrow A\times B\times C$ that satisfies sortlike conditions.

$g^{-1}\circ h:\left(A\times B\right)\times C\rightarrow A\times\left(B\times C\right)$ and $h^{-1}\circ g:A\times\left(B\times C\right)\rightarrow\left(A\times B\right)\times C$ are inverses of each other.

$g^{-1}\circ h$ is the natural isomorphism mentioned in the quote, but now in a general sense.

• I haven't been introduced to functors or bifunctors yet. Is there a means of doing this without their use? (Not to say that I don't now know what they are; I'm just assuming that the author of the book intended a differently styled solution). Commented Jul 24, 2015 at 16:52
• The quote in your question mentions "natural isomorphim". That is a concept that can only be properly introduced if functors are. Also in the quote you are asked to state and to prove. The statement is allready made in my answer so proving remains. It can be proved that in the situation as described there is a (natural) isomorphism $A\times(B\times C)\rightarrow (A\times B)\times C$. It can be found on base of the UMP (Unique Mapping Property) of products and involves the projections that go along with the products. Commented Jul 24, 2015 at 17:04
• I have edited something that might help. Commented Jul 24, 2015 at 17:39
• Aaaaaaaahhh, I tried so hard to get to that ismorphism!!! That is excellent. Thank you so much Commented Jul 24, 2015 at 17:53
• You are welcome. I am glad I could help. Commented Jul 25, 2015 at 7:20