Lemma 4.14.11. Let $M : \mathcal{I} \to \mathcal{C}$ be a diagram. Write $I = \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and $A = \text{Arrows}(\mathcal{I})$. Denote $s, t : A \to I$ the source and target maps. Suppose that $\prod _{i \in I} M_ i$ and $\prod _{a \in A} M_{t(a)}$ exist. Suppose that the equalizer of

$\xymatrix{ \prod _{i \in I} M_ i \ar@<1ex>[r]^\phi \ar@<-1ex>[r]_\psi & \prod _{a \in A} M_{t(a)} }$

exists, where the morphisms are determined by their components as follows: $p_ a \circ \psi = M(a) \circ p_{s(a)}$ and $p_ a \circ \phi = p_{t(a)}$. Then this equalizer is the limit of the diagram.

Proof. Omitted. $\square$

Comment #1021 by correction_bot on

In $p_a \circ \psi = a \circ p_{s(a)}$, the RHS should be $M(a) \circ p_{s(a)}$.

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