Direct limit of group When we study shaves we have that a germ is the direct limit of groups (set, vector spaces). But how can I show that the direct limit of groups is a group?
 A: The direct limit of a directed system of groups is always a group, because the transition maps are required to be group homomorphisms.
Suppose $\{G_i\}$, $i\in I$ and $\rho_{ij}:G_i\to G_j$ is a directed system of groups.  Define an operation on $\varinjlim G_i$ as follows:
$$\langle g,G_i\rangle\cdot\langle h,G_j\rangle=\langle\rho_{ik}(g)\cdot\rho_{jk}(h),G_k\rangle$$
where both $i\le k$ and $j\le k$.  Is the product well-defined?  Suppose $\langle g',G_{i'}\rangle=\langle g,G_i\rangle$ and $\langle h',G_{j'}\rangle=\langle h,G_j\rangle$, and perform the product of $\langle g',G_{i'}\rangle$ and $\langle h',G_{j'}\rangle$ in $G_{k'}$.  We can then find $l$ large enough so that both of $k,k'$ are less than $l$, $\rho_{i'l}(g')=\rho_{il}(g)$, and $\rho_{j'l}(h')=\rho_{jl}(h)$.  Then we have
$$\begin{align}\langle g',G_{i'}\rangle\cdot\langle h',G_{j'}\rangle&=\langle\rho_{i'k'}(g')\cdot\rho_{j'k'}(h'),G_{k'}\rangle\\&=\langle\rho_{k'l}(\rho_{i'k'}(g')\cdot\rho_{j'k'}(h')),G_{l}\rangle\\&=\langle\rho_{i'l}(g')\cdot\rho_{j'l}(h'),G_{l}\rangle\\&=\langle\rho_{il}(g)\cdot\rho_{jl}(h),G_{l}\rangle\\&=\langle\rho_{kl}(\rho_{ik}(g)\cdot\rho_{jk}(h)),G_{l}\rangle\\&=\langle\rho_{ik}(g)\cdot\rho_{jk}(h),G_{k}\rangle\\&=\langle g,G_{i}\rangle\cdot\langle h,G_{j}\rangle\end{align}$$
Notice where we used that $\rho_{k'l}$ and $\rho_{kl}$ are maps of groups.  So, in practice, group multiplication takes place in some $G_k$ for $k$ large enough.  With this well-defined product, all of the axioms for a group follow because they hold in $G_k$.
