Lemma 4.18.2. Let \mathcal{C} be a category. The following are equivalent:
Connected finite limits exist in \mathcal{C}.
Equalizers and fibre products exist in \mathcal{C}.
Lemma 4.18.2. Let \mathcal{C} be a category. The following are equivalent:
Connected finite limits exist in \mathcal{C}.
Equalizers and fibre products exist in \mathcal{C}.
Proof. Since equalizers and fibre products are finite connected limits we see that (1) implies (2). For the converse, let \mathcal{I} be a finite connected index category. Let F : \mathcal{J} \to \mathcal{I} be the functor of index categories constructed in the proof of Lemma 4.18.1. Then we see that we may replace \mathcal{I} by \mathcal{J}. The result is that we may assume that \mathop{\mathrm{Ob}}\nolimits (\mathcal{I}) = \{ x_1, \ldots , x_ n\} \amalg \{ y_1, \ldots , y_ m\} with n, m \geq 1 such that all nonidentity morphisms in \mathcal{I} are morphisms f : x_ i \to y_ j for some i and j.
Suppose that n > 1. Since \mathcal{I} is connected there exist indices i_1, i_2 and j_0 and morphisms a : x_{i_1} \to y_{j_0} and b : x_{i_2} \to y_{j_0}. Consider the category
with
and all other morphism sets the same as in \mathcal{I}. For any functor M : \mathcal{I} \to \mathcal{C} we can construct a functor M' : \mathcal{I}' \to \mathcal{C} by setting
and for a morphism f' : x \to y_ j corresponding to, say, f : x_{i_1} \to y_ j we set M'(f) = M(f) \circ \text{pr}_1. Then the functor M has a limit if and only if the functor M' has a limit (proof omitted). Hence by induction we reduce to the case n = 1.
If n = 1, then the limit of any M : \mathcal{I} \to \mathcal{C} is the successive equalizer of pairs of maps x_1 \to y_ j hence exists by assumption. \square
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Comment #6279 by Yuto Masamura on
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