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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 $i\!\leq\!j\!\leq\!k\!\in\!I$.

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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 $i\!\in\!I$.

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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 $\alpha'\!\circ\!\alpha'_i\!=\!\alpha_i$. enter image description here

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.

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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

1 Answer 1

up vote 3 down vote accepted

In the definition of a direct limit you say that in the given situation there is a unique morphism that completes the diagram.

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 $\alpha\circ\alpha'=1_{A'}$.

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"$1_A$ 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. –  Leon Sep 29 '11 at 10:10

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