Category has Finite Limits iff Finite Products and Equalizers
Theorem
Let $\mathbf C$ be a metacategory.
Then:
- $\mathbf C$ has all finite limits
- $\mathbf C$ 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:
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.
$\blacksquare$
Sources
- 2010: Steve Awodey: Category Theory (2nd ed.) ... (previous): $\S 5.4$: Proposition $5.21$