Bijection between direct limits 
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.
 A: 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.)
A: 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.
