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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