# uniqueness of a direct limit

DEFINITIONS: $$(I,\leq)$$ is a preordered set when $$I$$ is a set and $$\leq$$ is a reflexive and transitive binary relation on $$I$$, i.e. $$\forall i\!\in\!I\!: i\!\leq\!i$$ and $$\forall i,j,k\!\in\!I\!: i\!\leq\!j,j\!\leq\!k\Rightarrow i\!\leq\!k$$. An directed set is a preordered set $$I$$ such that $$\forall i,\!j\!\in\!I\,\exists k\!\in\!I\!:\,i,\!j\!\leq\!k$$.

Let $$I$$ be a directed preordered set and $$\underline{C}$$ a category. A direct system in $$\underline{C}$$ is a pair $$((A_i)_{i\in I},(\alpha_{i,j})_{i,j\in I,i\leq j})$$, denoted shorter with $$(A_i,\alpha_{i,j})_{i\leq j\in I}$$, that consists of a family $$A_i$$ of objects of $$\underline{C}$$ and a family $$\alpha_{i,j}$$ of morphisms of $$\underline{C}$$, such that: $$\alpha_{i,j}\!:A_i\!\rightarrow\!A_j$$ for $$i\!\leq\!j$$; and $$\alpha_{i,i}$$ is the identity morphism on $$A_i$$; and $$\alpha_{j,k}\!\circ\!\alpha_{i,j}\!=\!\alpha_{i,k}$$ for $$i\!\leq\!j\!\leq\!k\!\in\!I$$. In praxis, the most usual cases are $$\underline{C}$$ $$=$$ SET, GRP, RING, $$R$$-MOD, $$R$$-ALG, i.e. the category of sets, groups, rings, $$R$$-modules, $$R$$-algebras.

An object $$A$$ of $$\underline{C}$$, together with morphisms $$\alpha_i\!:A_i\!\rightarrow\!A$$ of $$\underline{C}$$, is a direct limit (or inductive limit, or directed colimit) of the direct system $$(A_i,\alpha_{i,j})_{i\leq j\in I}$$ in $$\underline{C}$$, denoted $$(A,\alpha_i)_{i\in I}\!=\!\varinjlim(A_i,\alpha_{i,j})_{i\leq j\in I}$$ or just $$A\!=\!\varinjlim(A_i,\alpha_{i,j})_{i\leq j\in I}$$ or $$A\!=\!\varinjlim(A_i)_{i\in I}$$, if $$\alpha_j\!\circ\!\alpha_{i,j}\!=\!\alpha_i$$ for $$i\!\leq\!j\!\in\!I$$; and the universal property is satisfied: for any other object $$A'$$ of $$\underline{C}$$ and morphisms $$\alpha'_i\!:A_i\!\rightarrow\!A'$$ of $$\underline{C}$$ satisfying $$\alpha'_j\!\circ\!\alpha_{i,j}\!=\!\alpha'_i$$, there exists a unique morphism $$\alpha\!:A\!\rightarrow\!A'$$ of $$\underline{C}$$ such that $$\alpha\!\circ\!\alpha_i\!=\!\alpha'_i$$ for all $$i\!\in\!I$$. QUESTION: how can I prove the statement:

"In any category, if a direct limit of a direct system exists, then it is unique up to isomorphism."

My attempt: If $$(A,\alpha_i)$$ and $$(A',\alpha'_i)$$ are both direct limits of $$(A_i,\alpha_{i,j})_{i\leq j\in I}$$, then $$\exists!\alpha\!:A\!\rightarrow\!A'$$ such that $$\alpha\!\circ\!\alpha_i\!=\!\alpha'_i$$, and $$\exists!\alpha'\!:A'\!\rightarrow\!A$$ such that $$\alpha'\!\circ\!\alpha'_i\!=\!\alpha_i$$. We wish to show $$\alpha'\circ\alpha=1_A$$ and $$\alpha\circ\alpha'=1_{A'}.$$

We have $$\alpha'\circ\alpha\circ\alpha_i=\alpha'\circ\alpha'_i=\alpha_i$$ and $$\alpha\circ\alpha'\circ\alpha'_i=\alpha\circ\alpha_i=\alpha'_i$$.

Also $$\alpha\circ\alpha'\circ\alpha=\alpha$$ and $$\alpha'\circ\alpha\circ\alpha'=\alpha'$$ by the uniqueness of the universal property.

This is where I run out of ideas.

• Hint: $\alpha'\alpha$ and $1_A$ are two morphisms $A \to A$ satisfying the same hypotheses granting their uniqueness. – t.b. Sep 29 '11 at 9:59

Now, when you want to prove that your two candidates $A$ and $A'$ for the direct limit are isomorphic, just observe that $1_A$ and $\alpha'\circ\alpha$ are both morphisms witnessing the universal property of $A$ with respect to $A$ itself.
Since such a morphism is unique, $\alpha'\circ\alpha=1_A$. The symmetric argument works for $\alpha\circ\alpha'=1_{A'}$.
• "$1_A$ and $α′\circ α$ are both morphisms witnessing the universal property of $A$ with respect to $A$ itself." Ah, that's the thing I didn't think of, thanks. – Leo Sep 29 '11 at 10:10