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In a paper that I'm writing, the following comes up:

Construction. Let $\mathscr C$ be a finitely complete category, and let $A_1 \to B_1, \ldots, A_n \to B_n$ be morphisms in $\mathscr C$. Let $B$ be an object with maps $\pi_i \colon B \to B_i$ for all $i \in \{1,\ldots, n\}$. Then construct the base change $$A = \bigg(\ldots\bigg(\bigg( B \underset{B_1}\times A_1\bigg) \underset{B_2}\times A_2 \bigg) \ldots \bigg) \underset{B_n}\times A_n$$ of $B$ along all morphisms $A_1 \to B_1, \ldots, A_n \to B_n$.

Picture. In the case $n = 2$, the picture looks as follows:

Picture for n=2

Neither square is a pullback, but $A$ is the limit of the rest of the diagram.

Question. Is there any alternative notation for the limit $A$ defined above?

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$A$ can be identified with the pullback $$B\times_{\prod_i B_i}\prod_i A_i. $$ (Which I think is nicer notation)

Proof:

Let $h : B\to \prod_i B_i$ be the canonical map. Let $k = \prod_i (A_i\to B_i)$ be the product of the individual maps, with the individual maps being $\alpha_i : A_i\to B_i$.

$$\newcommand\C{\mathscr{C}}\C\newcommand\of[1]{\left({#1}\right)}\of{X,B\times_{\prod_i B_i}\prod_i A_i} \simeq \{(f,g) \mid f:X\to B, g:X\to \prod_i A_i, hf = kg\}$$ For $g:X\to \prod_i A_i$, let $g_i$ be the $i$th component of $g$. Let $\pi_i : \prod_i B_i \to B_i$ be the projection maps.

Then $hf=kg$ if and only if $\pi_ihf = \pi_ikg$ for all $i$, but $\pi_i kg$ is equivalently the map $$ X\xrightarrow{g_i} A_i \xrightarrow{\alpha_i} B_i.$$ Thus $hf = kg$ if and only if $\pi_ihf = \alpha_i g_i$ for all $i$. Conversely, any family of maps $(g_i)_i$ with $\pi_ihf = \alpha_i g_i$ for all $i$ induces a map $g: X\to \prod A_i$ such that $hf=kg$. Hence

$$ \{(f,g) \mid f:X\to B, g:X\to \prod_i A_i, hf = kg\} \simeq $$ $$ \{(f,(g_i)_i) \mid f:X\to B, g_i:X\to A_i, \pi_ihf = \alpha_i g_i\} \simeq \C(X,A) $$ with the last natural isomorphism being by definition of the limit.

Thus by the Yoneda lemma, the claim follows. $\blacksquare$

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    $\begingroup$ Oh, of course! That's a big improvement. This is pretty close to how one constructs arbitrary finite limits from products and equalisers. $\endgroup$ – Remy Feb 4 '20 at 19:52
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    $\begingroup$ @Remy Yes, the intuition was the same, or at least similar. We replace a cone of maps with a single map to the product. $\endgroup$ – jgon Feb 5 '20 at 5:05

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