Category has Finite Limits iff Finite Products and Equalizers

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Theorem

Let $\mathbf C$ be a metacategory.


Then:

$\mathbf C$ has all finite limits

if and only if:

$\mathbf C$ has all finite products and equalizers.


Proof

Necessary Condition

By definition, finite products are instances of finite limits.

So are equalizers, by Equalizer as Limit.

$\Box$


Sufficient Condition

Let $C: \mathbf J \to \mathbf C$ be a diagram, with $\mathbf J$ finite.

The objective is to construct the limit $D = \varprojlim_j C_j$.

That is:

an object $D$
morphisms $p_j: D \to C_j$

such that:

$C_\alpha p_i = p_j$ for each $\mathbf J$-morphism $\alpha: i \to j$
for each $E$ and $q_j: E \to C_j$ such that $C_\alpha q_i = q_j$ for each $\alpha$, there is a unique $f: E \to D$ such that $q_j = p_j f$ for each $j$


To this end, consider the finite products:

$\ds \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ and $\ds \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$

of the objects of the diagram and the codomains of its morphisms.

Next, define the morphisms $\pi, \varepsilon: \ds \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j \to \prod_{\alpha \mathop \in \operatorname{mor} \mathbf J} C_{\operatorname{cod} \alpha}$ by:

$\pi_\alpha = \pr_{C_{\operatorname{cod} \alpha}}$
$\varepsilon_\alpha = C_\alpha \pr_{C_{\operatorname{dom} \alpha}}$

Now let $e: D \to \ds \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$ be the equalizer of $\pi$ and $\varepsilon$.

Also, define $p_j: D \to C_j$ by $p_j = \pr_{C_j} e$ for each $j$.


Now, for every $\alpha: i \to j$:

\(\ds C_\alpha p_i\) \(=\) \(\ds C_\alpha \pr_{C_j} e\) Definition of $p_i$
\(\ds \) \(=\) \(\ds \varepsilon_\alpha e\) Definition of $\varepsilon_\alpha$
\(\ds \) \(=\) \(\ds \pi_\alpha e\) $e$ equalizes $\pi$ and $\varepsilon$
\(\ds \) \(=\) \(\ds \pr_{C_j} e\) Definition of $\pi_\alpha$
\(\ds \) \(=\) \(\ds p_j\) Definition of $p_j$


Lastly, suppose that $E$ and $q_j$ have the properties stated.

By definition of the product $\ds \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$, there is a morphism:

$\ds q: E \to \prod_{j \mathop \in \operatorname{ob} \mathbf J} C_j$

Now, for every $\alpha: i \to j$:

\(\ds \pi_\alpha q\) \(=\) \(\ds \pr_j q\) Definition of $\pi_\alpha$
\(\ds \) \(=\) \(\ds q_j\) Definition of $q$
\(\ds \) \(=\) \(\ds C_\alpha q_i\) Assumption on the $q_i$
\(\ds \) \(=\) \(\ds C_\alpha \pr_i q\) Definition of $q$
\(\ds \) \(=\) \(\ds \varepsilon_\alpha q\) Definition of $\varepsilon_\alpha$

Hence, since $e$ equalizes $\pi$ and $\varepsilon$, it follows that there exists a unique $f: E \to D$, such that:

$q = ef$

Thus, for each $j$:

$\pr_j q = \pr_j e f$

which, by the definitions of $q_j$ and $p_j$, amounts to:

$q_j = p_j f$


Hence $D$ and the $q_j$ form the desired limit.

$\blacksquare$


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