I am trying to show that if $G_{\alpha}$ form a directed set of torsionfree abelian groups, then $\varinjlim G_{\alpha}$ is torsionfree abelian.
One approach I thought of is to use a Zorn's lemma argument on the union of a chain of torsionfree abelian groups. An infinite direct product also works if the system is countable?
Another more worked out approach is as follows: Any $g \in \varinjlim G_{\alpha}$ has the form $\mu_{\alpha} (g_{\alpha})$ so that, if $n * g = 0$, then $0 = n*g = n * \mu_{\alpha} (g_{\alpha}) = \mu_{\alpha} (n * g_{\alpha})$. I want to say that $n * g_{\alpha}$ must then be 0 in order to reach a contradiction.
As you are working in a directed set, each element of the direct limit is a suitable equivalence class $[g,G_a]$ where $g\in G_a$. Also $[g,G_a]=0$ iff $\phi_{ab}(G_a)=0$ in $G_b$ for some $b\ge a$ in the index set $A$. For the direct limit to have torsion, then there must be some $n\in\Bbb N$ and a $[g,G_a]\ne0$ with $[ng,G_a]=0$. Then there is $b\ge a$ with $\phi_{ab}(ng)=0$ in $G_b$, that is, $n\phi_{ab}(g)=0$ in $G_b$. But $\phi_{ab}(g)=0$ as $G_b$ is torsion-free. That means that $[g,G_a]=0$, contradiction.