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The same question is given here Categorical products are associative, however, one of the answers uses concepts not introduced yet in my book, and the other provides a link (https://compose.ioc.ee/categoryTheory2020/week3/week3.pdf, "Diagram Chasing") which to me is still unclear. I am trying to prove $A \times (B \times C) \cong (A \times B) \times C$

Basically, the answer in the link states that we can form the following four commuting diagrams:

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From these we have the diagram:

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Then the author concludes that because this arrow is unique, it must be the identity arrow. This is the part that I do not follow. Where in the proof have we concluded that $q_2 \circ q_1$ makes the last diagram commute? Once that is established then I understand how we can conclude $q_2 \circ q_1 = 1_{A(BC)}$.

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1 Answer 1

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The commutativity of the last diagram follows from the commutativities of the previous four diagrams.

For the left-hand triangle we have $$ p_{A2} ∘ q_2 ∘ q_1 = p_{A1} ∘ p_{AB} ∘ q_1 = p_{A1} ∘ q_{AB} = p_{A2} $$ by the definition of $q_2$, the definition of $q_2$, and the definition of $q_{AB}$.

For the right-hand triangle we have \begin{align*} p_{BC} ∘ q_2 ∘ q_1 &= q_{BC} ∘ q_1 \tag{$1$} \\ &= ⟨p_{B1} ∘ p_{AB}, p_{C1}⟩ ∘ q_1 \tag{$2$} \\ &= ⟨p_{B1} ∘ p_{AB} ∘ q_1, p_{C1} ∘ q_1⟩ \\ &= ⟨p_{B1} ∘ q_{AB}, p_{C2} ∘ p_{BC}⟩ \tag{$3$} \\ &= ⟨p_{B2} ∘ p_{BC}, p_{C2} ∘ p_{BC}⟩ \tag{$4$} \\ &= ⟨p_{B2}, p_{C2}⟩ ∘ p_{BC} \\ &= 1_{BC} ∘ p_{BC} \\ &= p_{BC} \,, \end{align*} where for $(1)$ we use the definition of $q_2$, for $(2)$ we use the definition of $q_{BC}$, for $(3)$ we use the definition of $q_1$, for $(4)$ we use the definition of $q_{AB}$.

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