Flip map on a direct limit of tensor products Let $$G_1\xrightarrow{f_1} G_2\xrightarrow{f_2}G_2\to\cdots$$ be an direct system of abelian groups with direct limit $(G,\{f_{n,\infty}\}_n)$. 
For every $n\in\mathbb{N}$, let $\tau_n:G_n\otimes _\mathbb{Z}G_n\to G_n\otimes _\mathbb{Z}G_n$ given by $\tau_n(g\otimes h)=h\otimes g$ on elementary tensors. 
Writing $G=\varinjlim G_n$, does $\tau_n$ induce a "flip"map $\tau:G\otimes G\to G\otimes G$, $g\otimes h\mapsto h\otimes g$? How to verify that $\tau_n$ extends to such a flip-map on direct limits?
I tried to check this as follows: At first I considered $(f_{n,\infty}\otimes f_{n,\infty})\circ \tau_n$ and to check that this map extends to the direct limits, one has to check that $f_{n-1,\infty}\otimes f_{n-1,\infty}=(f_{n,\infty}\otimes f_{n,\infty})\circ \tau_n\circ (f_{n-1,n}\otimes f_{n-1,n})$ for every n. But I got that this equation isn't true. And I don't have any other ideas. So, how to do this?
 A: The equation you need to check is not the one you wrote but instead $$(f_{n-1,\infty}\otimes f_{n-1,\infty})\color{red}{\circ\tau_{n-1}}=(f_{n,\infty}\otimes f_{n,\infty})\circ \tau_n\circ (f_{n-1,n}\otimes f_{n-1,n}).$$  This guarantees that you get a well-defined map $\tau:G\otimes G\to G\otimes G$ by defining $\tau((f_{n,\infty}\otimes f_{n,\infty})(x))=(f_{n,\infty}\otimes f_{n,\infty})(\tau_n(x))$ whenever $x\in G_n\otimes G_n$ (since $G\otimes G$ is the direct limit of the system formed by the $G_n\otimes G_n$).  Checking this equation is straightforward: since $G_{n-1}\otimes G_{n-1}$ is generated by elements of the form $a\otimes b$, it suffices to check it when each side is evaluated on such an element, and both sides evaluate to $f_{n-1,\infty}(b)\otimes f_{n-1,\infty}(a)$ (the right-hand side does so since $f_{n,\infty}\circ f_{n-1,n}=f_{n-1,\infty})$.
This gives you a map $\tau:G\otimes G\to G\otimes G$.  To check that it is the flip map, it suffices to check that $\tau(a\otimes b)=b\otimes a$ for any $a,b\in G$.  But for any $a,b\in G$ there is some $n$ and $c,d\in G_n$ such that $a=f_{n,\infty}(c)$ and $b=f_{n,\infty}(d)$, and we have $$\tau(a\otimes b)=\tau((f_{n,\infty}\otimes f_{n,\infty})(c\otimes d))=(f_{n,\infty}\otimes f_{n,\infty})(\tau_n(c\otimes d))=(f_{n,\infty}\otimes f_{n,\infty})(d\otimes c)=b\otimes a.$$
