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Let $I$ be a directed poset. Let $J \subset I$ be a final subset, which means that $\forall i \in I\ \exists j \in J : i \le j$. Let $\left\{A_i \right\}_{i \in I} $ be a direct system of sets. Construct a natural bijection between direct limits $\lim _{ \ i \in I}A_i$ and $\lim_{ \ j \in J}A_j$.

I've read about direct limits in Grillet's Abstract Algebra, but I still don't know how to prove that.

Could you help me with that?

Thank you.

Bilbo
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  • A dircted system consists not only of sets $A_i$ but also maps (in other categories: morphisms) $\alpha_{i,j}: A_i\to A_j$ such that $\alpha_{j,k}\circ\alpha_{i,j}=\alpha_{i,k}$ for all $i\le j\le k$. A direct limit is then defined as an object $A$ together with morphisms $\alpha_i:A_i\to A$ with the universal property that for all $f_i:A_i\to B$ with $f_i=f_J\circ \alpha_{i,j}$ there is a unique $f:A\to B$ such that $f_i=f\circ\alpha_i$. Then check that every direct limits of $(A_i){i\in I}$ is a direct limit for $(A_i){i\in J}$ (and vice versa) if $J$ is cofinal in $I$. – Jochen Mar 19 '13 at 07:41
  • If $J$ is cofinal in $I$, then $\forall i\in I \ \exists j \in J: \ j \ge i$. If there is a mapping $(\lambda_i){i \in I}: (A_i){i\in I} \rightarrow L$, then $L$ is a direct limit of $(A_i){i\in I}$ if for every family of mappings $(\varphi_i){i\in I}: (A_i)_{i\in I} \rightarrow B$ there exists a unique mapping $\varphi_1: L \rightarrow B$ such that $\varphi = \varphi_1 \ \lambda$ – Bilbo Mar 19 '13 at 08:12
  • If for every element of $I$ there exists a bigger element in $J$, the relation between direct limits seems obvious because the arrows representing mappings in direct systems go from sets with smaller indices to sets with bigger indices. But what about the bijection? Could you give some more help? – Bilbo Mar 19 '13 at 08:22

2 Answers2

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This is a straightforward diagram chase: you can use the universal property of direct limits to show that there is a natural map from the limit over $I$ to the limit over $J$ and vice-versa, and then use it again to show that these maps are inverses of each other.

Let the directed system on $<I,\le>$ have objects $A_i$ ($i\in I$) and maps $\alpha_{i,k}: A_i\to A_k$ ($i\le k\in I$), let the direct limit $L:=\lim_{i\in I} A_i$ have maps $\alpha_i:A_i\to L$ ($i\in I$), let $J$ be cofinal in $I$, and let the direct limit $M:=\lim_{j\in J} A_j$ have maps $\beta_j:A_j\to M$ ($j\in J$.) By the cofinality of $J$, for each $i\in I$, you can pick some $q(i)\in J$ such that $i\le q(i)$; in fact, you can let $q(j)=j$ if $j\in J$. Also, since $I$ is directed and by the cofinality of $J$, for each pair of elements $i, k\in I$, you can pick some $r(i,k)\in J$ such that $q(i)\le r(i,k)$ and $q(k)\le r(i,k)$.

Now (writing function composition as juxtaposition), if $i, k\in I$ satisfy $i\le k$, $$ \alpha_{q(i),r(i,k)} \alpha_{i,q(i)}=\alpha_{i,r(i,k)}=\alpha_{q(k),r(i,k)} \alpha_{k,q(k)}\alpha_{i,k} $$ so, by the above, and since $M$ is a direct limit over $J$, $$ \beta_{q(i)} \alpha_{i,q(i)}=\beta_{r(i,k)} \alpha_{q(i),r(i,k)}\alpha_{i,q(i)} = \beta_{r(i,k)}\alpha_{q(k),r(i,k)} \alpha_{k,q(k)}\alpha_{i,k} = \beta_{q(k)} \alpha_{k,q(k)}\alpha_{i,k}. $$ Therefore, by the existence part of the universal property of $L$, you can find $\Phi:L\to M$ such that $$ \Phi \alpha_i = \beta_{q(i)} \alpha_{i,q(i)}, \qquad \ \ \ \text{for all } i\in I. $$ The other direction is simpler as, since $L$ is a direct limit over $I$, you immediately have $$ \alpha_\ell \alpha_{j,\ell} = \alpha_j , \qquad \ \ \ \text{for all } j, \ell\in J,\ j\le \ell, $$ so, by the existence part of the universal property of $M$, you can find $\Psi: M\to L$ such that $$ \Psi \beta_j = \alpha_j, \qquad \ \ \ \text{for all } j\in J. $$ Now, for each $j\in J$, $$ (\Phi \Psi) \beta_j = \Phi \alpha_j = \beta_{q(j)} \alpha_{j,q(j)}=\beta_j \alpha_{j,j}=\beta_j=1_{M} \beta_j, $$ so, by the uniqueness part of the universal property of $M$, $\Phi\Psi=1_M$. Similarly, for each $i\in I$, $$ (\Psi\Phi) \alpha_i = \Psi \beta_{q(i)} \alpha_{i,q(i)}=\alpha_{q(i)} \alpha_{i,q(i)}=\alpha_i = 1_L \alpha_i,$$ so, by the uniqueness part of the universal property of $L$, $\Psi\Phi=1_L$. Therefore, $\Phi$ is an isomorphism from $L$ to $M$ (and, if isomorphisms are bijections in the category you are working over, therefore also a bijection.)

David Moews
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The proof can be found in any complete introduction to category theory, for example Theorem IX.2.1 in Mac Lane's Categories for the working mathematician.

Martin Brandenburg
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