Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3

Lemma
\[ h_\infty: \begin{aligned} G_\infty &\longrightarrow H \\ \left[\!\left[{\left({x_n, n}\right)}\right]\!\right] &\longmapsto h_n \left({x_n}\right) \end{aligned} \] is a well-defined group homomorphism.

Proof
Once we have proven that this is well-defined it follows immediately that $h_\infty$ is a homomorphism, since the multiplication reduces to that on one specific $G_l \in \left( G_n \right)_{n \in \mathbb{N}}$, and $h_l$ is a homomorphism.

Thus let $(x_n,n),(x_{n'},n') \in \left[\!\left[{\left({x_n, n}\right)}\right]\!\right]$. W.l.o.g. let $n' \geqslant n$. Then we have $g_{nn'}(x_n) = x_{n'}$ \[		h_{n'} (x_{n'}) = h_{n'} \left( g_{nn'}(x_n) \right) = \underbrace{\left( h_{n'} \circ g_{nn'} \right)}_{= h_n}(x_n)	= h_n(x_n) \]		which proves that $h_\infty$ is independent of the representative chosen, i.e. well-defined.