Category has Finite Limits iff Finite Products and Equalizers

Theorem
Let $\mathbf C$ be a metacategory.

Then:


 * $\mathbf C$ has all finite limits

iff:


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

Necessary Condition
By definition, finite products are instances of finite limits.

So are equalizers, by Equalizer as Limit.

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:


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

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

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


 * $\pi_\alpha = \operatorname{pr}_{C_{\operatorname{cod} \alpha}}$
 * $\varepsilon_\alpha = C_\alpha \operatorname{pr}_{C_{\operatorname{dom} \alpha}}$

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

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

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

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

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


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

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

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 $j$:


 * $\operatorname{pr}_j q = \operatorname{pr_j} e f$

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.