## 4.18 Finite limits and colimits

A finite (co)limit is a (co)limit whose index category is finite, i.e., the index category has finitely many objects and finitely many morphisms. A (co)limit is called nonempty if the index category is nonempty. A (co)limit is called connected if the index category is connected, see Definition 4.16.1. It turns out that there are “enough” finite index categories.

Lemma 4.18.1. Let $\mathcal{I}$ be a category with

1. $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ is finite, and

2. there exist finitely many morphisms $f_1, \ldots , f_ m \in \text{Arrows}(\mathcal{I})$ such that every morphism of $\mathcal{I}$ is a composition $f_{j_1} \circ f_{j_2} \circ \ldots \circ f_{j_ k}$.

Then there exists a functor $F : \mathcal{J} \to \mathcal{I}$ such that

1. $\mathcal{J}$ is a finite category, and

2. for any diagram $M : \mathcal{I} \to \mathcal{C}$ the (co)limit of $M$ over $\mathcal{I}$ exists if and only if the (co)limit of $M \circ F$ over $\mathcal{J}$ exists and in this case the (co)limits are canonically isomorphic.

Moreover, $\mathcal{J}$ is connected (resp. nonempty) if and only if $\mathcal{I}$ is so.

Proof. Say $\mathop{\mathrm{Ob}}\nolimits (\mathcal{I}) = \{ x_1, \ldots , x_ n\}$. Denote $s, t : \{ 1, \ldots , m\} \to \{ 1, \ldots , n\}$ the functions such that $f_ j : x_{s(j)} \to x_{t(j)}$. We set $\mathop{\mathrm{Ob}}\nolimits (\mathcal{J}) = \{ y_1, \ldots , y_ n, z_1, \ldots , z_ n\}$ Besides the identity morphisms we introduce morphisms $g_ j : y_{s(j)} \to z_{t(j)}$, $j = 1, \ldots , m$ and morphisms $h_ i : y_ i \to z_ i$, $i = 1, \ldots , n$. Since all of the nonidentity morphisms in $\mathcal{J}$ go from a $y$ to a $z$ there are no compositions to define and no associativities to check. Set $F(y_ i) = F(z_ i) = x_ i$. Set $F(g_ j) = f_ j$ and $F(h_ i) = \text{id}_{x_ i}$. It is clear that $F$ is a functor. It is clear that $\mathcal{J}$ is finite. It is clear that $\mathcal{J}$ is connected, resp. nonempty if and only if $\mathcal{I}$ is so.

Let $M : \mathcal{I} \to \mathcal{C}$ be a diagram. Consider an object $W$ of $\mathcal{C}$ and morphisms $q_ i : W \to M(x_ i)$ as in Definition 4.14.1. Then by taking $q_ i : W \to M(F(y_ i)) = M(F(z_ i)) = M(x_ i)$ we obtain a family of maps as in Definition 4.14.1 for the diagram $M \circ F$. Conversely, suppose we are given maps $qy_ i : W \to M(F(y_ i))$ and $qz_ i : W \to M(F(z_ i))$ as in Definition 4.14.1 for the diagram $M \circ F$. Since

$M(F(h_ i)) = \text{id} : M(F(y_ i)) = M(x_ i) \longrightarrow M(x_ i) = M(F(z_ i))$

we conclude that $qy_ i = qz_ i$ for all $i$. Set $q_ i$ equal to this common value. The compatibility of $q_{s(j)} = qy_{s(j)}$ and $q_{t(j)} = qz_{t(j)}$ with the morphism $M(f_ j)$ guarantees that the family $q_ i$ is compatible with all morphisms in $\mathcal{I}$ as by assumption every such morphism is a composition of the morphisms $f_ j$. Thus we have found a canonical bijection

$\mathop{\mathrm{lim}}\nolimits _{B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{J})} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M(F(B))) = \mathop{\mathrm{lim}}\nolimits _{A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})} \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(W, M(A))$

which implies the statement on limits in the lemma. The statement on colimits is proved in the same way (proof omitted). $\square$

Lemma 4.18.2. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Connected finite limits exist in $\mathcal{C}$.

2. 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

$\mathcal{I}' = \{ x\} \amalg \{ x_1, \ldots , \hat x_{i_1}, \ldots , \hat x_{i_2}, \ldots x_ n\} \amalg \{ y_1, \ldots , y_ m\}$

with

$\mathop{\mathrm{Mor}}\nolimits _{\mathcal{I}'}(x, y_ j) = \mathop{\mathrm{Mor}}\nolimits _\mathcal {I}(x_{i_1}, y_ j) \amalg \mathop{\mathrm{Mor}}\nolimits _\mathcal {I}(x_{i_2}, y_ j)$

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

$M'(x) = M(x_{i_1}) \times _{M(a), M(y_{j_0}), M(b)} M(x_{i_2})$

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. $\square$

Lemma 4.18.3. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Nonempty finite limits exist in $\mathcal{C}$.

2. Products of pairs and equalizers exist in $\mathcal{C}$.

3. Products of pairs and fibre products exist in $\mathcal{C}$.

Proof. Since products of pairs, fibre products, and equalizers are limits with nonempty index categories we see that (1) implies both (2) and (3). Assume (2). Then finite nonempty products and equalizers exist. Hence by Lemma 4.14.11 we see that finite nonempty limits exist, i.e., (1) holds. Assume (3). If $a, b : A \to B$ are morphisms of $\mathcal{C}$, then the equalizer of $a, b$ is

$(A \times _{a, B, b} A)\times _{(\text{pr}_1, \text{pr}_2), A \times A, \Delta } A.$

Thus (3) implies (2), and the lemma is proved. $\square$

Lemma 4.18.4. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Finite limits exist in $\mathcal{C}$.

2. Finite products and equalizers exist.

3. The category has a final object and fibre products exist.

Proof. Since finite products, fibre products, equalizers, and final objects are limits over finite index categories we see that (1) implies both (2) and (3). By Lemma 4.14.11 above we see that (2) implies (1). Assume (3). Note that the product $A \times B$ is the fibre product over the final object. If $a, b : A \to B$ are morphisms of $\mathcal{C}$, then the equalizer of $a, b$ is

$(A \times _{a, B, b} A)\times _{(\text{pr}_1, \text{pr}_2), A \times A, \Delta } A.$

Thus (3) implies (2) and the lemma is proved. $\square$

Lemma 4.18.5. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Connected finite colimits exist in $\mathcal{C}$.

2. Coequalizers and pushouts exist in $\mathcal{C}$.

Proof. Omitted. Hint: This is dual to Lemma 4.18.2. $\square$

Lemma 4.18.6. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Nonempty finite colimits exist in $\mathcal{C}$.

2. Coproducts of pairs and coequalizers exist in $\mathcal{C}$.

3. Coproducts of pairs and pushouts exist in $\mathcal{C}$.

Proof. Omitted. Hint: This is the dual of Lemma 4.18.3. $\square$

Lemma 4.18.7. Let $\mathcal{C}$ be a category. The following are equivalent:

1. Finite colimits exist in $\mathcal{C}$.

2. Finite coproducts and coequalizers exist in $\mathcal{C}$.

3. The category has an initial object and pushouts exist.

Proof. Omitted. Hint: This is dual to Lemma 4.18.4. $\square$

Comment #3992 by Kazuki Masugi on

There is a typo in the proof of Lemma 04AT.,"$\times _{M(a),M(y_j),M(b)}$" will be "$\times_{M(a),M(y_{j_0}),M(b)}$".

Comment #3993 by Kazuki Masugi on

There is a typo in the proof of Lemma 04AT.,"$\times _{M(a),M(y_j),M(b)}$" will be "$\times_{M(a),M(y_{j_0}),M(b)}$".

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