Question

Assume $C$ is a locally small category with equalizers in the sense that given any two arrows $f$ and $g$ with a common source $a$ and target $b$, then there is an object $e$ and an arrow $i\colon e\to a$ such that $fi = gi$ that satisfies the usual universal property. Seems that this generalizes fairly easily (by induction) to any finite subset of the hom-set $C(a,b)$. That is, given a finite $A\subset C(a,b)$, there is an arrow $i\colon e\to a$ such that for any $f,g\in A$, we have $fi=gi$ and if $k\colon c\to a$ is such that for any $f,g\in A$ we have $fk=gk$, then there is a unique $h\colon c\to e$ such that $k=ih$.

My question is whether this can be generalized to all of $C(a,b)$ when the hom-set is not necessarily finite.

Answer 1

Look for "equalizer of a family of arrows" in an arbitrary book on category theory. It is a special case of a limit. A typical example for a category which has finite limits, but not arbitrary limits (and therefore not all infinite equalizers), is the category of schemes.

Answer 2

Note that you were using induction to show the existence of equilizers for a finite sets of morphisms in $\mathrm{Hom}(A,B)$. So one may be thinking of using transfinite induction to access any subset of maps. There is a way of doing this without asking for completeness. You just require that all the ordinal limits exist. By this I mean, that for any ordinal $\lambda$ the limit of any $\lambda$-diagram exists ($\mathrm{Ob}\lambda=\lambda$ and there exists a unique morphism from $a$ to $b$ if $a\geq b$).

One can also think of this business as requiring the existence of limits of directed systems (I think it is equivalent to requiring that ordinal limits exist). In this case if we are given a subset $S\subset\mathrm{Hom}(A,B)$, let $F(S)$ denote the category of all of its finite subsets under inclusion. This is a directed system. For each $D\in F(S)$ we know there exists a coequalizer, and this can be extended to a cofunctor $F(S)$ to our category. The limit of this diagram is going to equalize $S$.

Answer 3

Let $D$ the graph with objects $i(a)$ and $i(f),\ f\in C(a, b)$ and arrows $m(f): i(a)\to i(f)$. Then if $F: D\to C$ is the diagam $i(a)\mapsto a$, $i(f)\mapsto b$, $m(f)\mapsto f$ the limit o $F$ is the equalizer af all elments of $C(a, b)$. If $C$ has a zero object $0$ (i.e. 0 is initial and final) and has these generalized kernel, then $C$ has product (let $a=0$), and of course kernels of pairs, then $C$ is neccesarily complete.

Edit: A example (in addition to that of Andreas Blass) of a category with "book"-limit (on diagrams like above) that isnt complete (i.e. whitout products).

Let $Sem\subset Set$ the subcategory of sets and inective maps, then let $A,\ B$ two finite set with more than one elements, and let $1$ the final (one element set) object. Of course for any set $X$ we have $Sem(1, X)= Set(1, X)=|X|$ (cardinality of $X$) then if the product of $A, B$ in $Sem$ exist (indicate it by $A\times_m B$) its cardinality is:

$|A\times_m B|= |Set(1, A\times_m B)|= |Sem(1, A\times_m B)|=|Sem(1, A)\times Sem(1, B)|=$

$|Set(1, A)\times Set(1, B)|= |A\times B| $

then $A\times_m B$ cannot exist because the natural projection $\pi: A\times_m B\to A$ cannot be injective.

Answer 4

Consider the set of natural numbers as a monoid with the operation of "maximum" (so 0 is the identity element), and view it as a category with one object $*$. (That is, the morphisms from $*$ to $*$ are the natural numbers, and the composition of two of these is the larger of the two.) Then the equalizer of any two of morphisms $a,b$ exists; it is (the object $*$ with) the morphism $\max\{a,b\}$. (There is no significance to the fact that this is also the composite of $a$ and $b$.) But no morphism equalizes all the natural numbers --- specifically, any $a$ fails to equalize $a$ and $a+1$.