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"for some $f' \in R$" should be "for some $f'' \in S$".

And here the definition of partial ordered set is not required to have "antisymetry" (it is the definition tagged 00D3), while the definition of partial ordered set in wiki page is required to have the antisymetry property.

@#588: Hmm... I see what you mean. We chose Definition 4.21.1 because it works well for directed limits, etc. But perhaps we should change it... not sure yet.

@#587: OK, I fixed this by requiring $f'' \in R$ which is good enough to define the map. See here.

@#588: This is now addressed here.

About the hint: one should first check that the map $M_{f'}\to M_f$ is well defined, i.e. independent of $f''$. Of course this can also be done via the universal property (even avoiding the use of $f''$), so this property is applied "twice".

@#6485. OK, I am going to leave it as is and hope that confused readers will find your comment and benefit from it.

Hint: Don't try to use Lemma 10.9.7 to prove this; the condition that $S$ should act as automorphisms on $N$ is not satisfied in this case. Instead, just prove it directly from the definitions of colimit and localization.

Proof: Let $N$ be an $R$-module and $\phi_f: M_f\to N$ be any set of homomorphisms commuting with the given maps $\varphi_{f',f}:M_{f'} \to M_f$. The goal is to show that there exists a unique homomorphism $\psi: S^{-1}M\to N$ such that $\psi(m/s) = \phi_s(m/s)$ for all $m\in M$ and $s\in S$. The requirement $\psi(m/s) = \phi_s(m/s)$ automatically implies uniqueness, so we need only show that $\psi$ is well-defined and $R$-linear. Key to showing these two properties is the following result: for any $t\in S$ we have $$\phi_s(m/s) = \phi_{st}(tm/ts)$$ becuase $\phi_s(m/s) = \phi_{st}\circ \varphi_{s,st}(m/s) = \phi_{st}(tm/ts)$. To show that $\psi$ is well-defined, suppose $m/s = m'/t$. Let $u\in S$ be an element satisfiying the equvalent condition $u(tm-sm') = 0$. Then we have $$\psi(m/s) = \phi_s(m/s) = \phi_{uts}(utm/uts) = \phi_{ust}(usm'/ust) = \phi_t(m'/t) = \psi(m'/t)$$ hence $\psi$ is well-defined. To show $R$-linearity is straight forward and left as an exercise.

Comment on the notation in the lemma: Please change the exponent from $e$ to $n$, as currently it is possible to wonder whether the $e$ has some special meaning, like the $e$ used in an extended ideal.