CatDat

Construction of a colimit of a sequence of monomorphisms as a quotient of a congruence

Lemma.

Let C\C be a countably extensive category with quotients of congruences. Then C\C has colimits of sequences of monomorphisms.

Proof. Suppose we have a sequence X1X2X_1 \hookrightarrow X_2 \hookrightarrow \cdots with corresponding monomorphisms fm,n:XmXnf_{m,n} : X_m \hookrightarrow X_n for mnm \le n. Define YY to be the coproduct of all XnX_n. Now for each mnm\le n, define Em,nXmE_{m,n} \coloneqq X_m with two maps im,infm,n:Em,nYi_m, i_n \circ f_{m,n} : E_{m,n} \rightrightarrows Y, and similarly for mnm \ge n define Em,nXnE_{m,n} \coloneqq X_n with two maps imfn,m,in:Em,nYi_m \circ f_{n,m}, i_n : E_{m,n} \rightrightarrows Y. Then the coproduct of all Em,nE_{m,n}, with the induced morphisms to YY, forms a congruence. Here to prove the maps are jointly monomorphic, and again in proving transitivity, we use extensivity to split the domain of the generalized elements of m,n=1Em,n\sum_{m,n=1}^\infty E_{m,n} so that without loss of generality we may assume each factors through one of the coproduct inclusions. Now a quotient of this congruence must be a colimit of the sequence. \square

Author: Daniel Schepler

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