Direct limit of a series of maps which are left multiplication by $f$ Let $M$ be a module over a ring $R$, and let $f\in R$.
Then the direct limit of the diagram $$M\rightarrow M\rightarrow M\rightarrow M\rightarrow \cdots,$$ where each map is given by $m\mapsto fm$ is the module $M[f^{-1}] = \{\frac{m}{f^n} : m\in M, n\in\mathbb N\}$.
I can see why $M[f^{-1}]$ satisfies the property which makes the diagram commute. At each stage we take the map $\phi_k : M\rightarrow M[f^{-1}]$ which maps $m\mapsto \frac{m}{f^k}$. But I can't see why it satisfies the universal property.
Any help would be greatly appreciated.
 A: Just start working through the universal property. (As an alternate proof, you can consider the universal property of localization.)
Given a module $N$ and a family of homomorphisms $\psi_k : M \to N$ such that $\psi_{k+1}(fm) = \psi_k(m)$, we need to construct a homomorphism $h_\psi : M[f^{-1}] \to N$ such that $\psi_k = h_\psi\circ\phi_k$ which is to say $h_\psi(\frac{m}{f^k}) \equiv \psi_k(m)$ which is well-defined because $h_\psi(\frac{m}{f^k}) = h_\psi(\frac{fm}{f^{k+1}}) = \psi_{k+1}(fm) = \psi_k(m)$. So such an $h_\psi$ exists and is unique. There's a naturality condition to check which turns out to be trivially true once you formulate it.
In different language, you're trying to prove that $_R\mathbf{Mod}(M[f^{-1}],-)\cong [\mathbb N,{}_R\mathbf{Mod}](F,\Delta -)$ where $\mathbb N$ is the natural numbers with their standard ordering viewed as a category, $[\mathbb N,{}_R\mathbf{Mod}]$ is the category of functors $\mathbb N \to {}_R\mathbf{Mod}$, $\Delta : {}_R\mathbf{Mod}\to[\mathbb N,{}_R\mathbf{Mod}]$ is the diagonal functor (i.e. it produces a constant functor), and $F$ is the functor such that $F(n) = M$ and $F(n < n+1) = f$. This last is just the diagram you described at the beginning. $\phi \in [\mathbb N,{}_R\mathbf{Mod}](F,\Delta M[f^{-1}])$, $\psi \in [\mathbb N,{}_R\mathbf{Mod}](F,\Delta N)$, and $h_\psi$ is the arrow corresponding to $\psi$ by the above natural isomorphism.  $h_\phi$ should be $id$, which it is.
