Condition for isomorphism of $R$-module homomorphism. In the proof of theorem 9.4.7 of Charles Weibel's "An Introduction to Homological Algebra" it is claimed to aid the proof that an $R$-module homomorphism $\psi$ is an "isomorphism if and only if $\psi\otimes_RR_m$ is an isomorphism for every maximal ideal $m$" where $R_m$ is intended to be the localisation of $R$ at $m$.
The result is also used in a similar proof by Jean-Louis Loday's "Cyclic Homology" (page 102).
As far as I can tell this is not something proven in the sources and I couldn't find the result elsewhere or come up with a proof myself. Is there an obvious reason why this should be true? Where can I find or what would a proof of such a statement look like?
 A: The following is from Atiyah and Macdonald's book on commutative algebra (in chapter 3). First, localization is an exact operation, which gives one half of the theorem. To see the other half, we only need the fact that for an $R$-module $M,$ $M_{\mathfrak m}=0$ for every maximal ideal $\mathfrak m$ of $R$ implies that $M=0.$ The proof is like this.
Suppose $M\neq 0$ and let $0\neq x\in M.$ Then $\mathrm{Ann}(x)$ is a proper ideal of $R$ and so is contained in some maximal ideal $\mathfrak m.$ If the image of $x$ in $M_\mathfrak m$ were zero, then $x$ would be killed by some element of $R-\mathfrak m,$ which is impossible since $\mathrm{Ann}(x)\subset \mathfrak m.\square$
Now that we have this, if $\psi_\mathfrak m:M_\mathfrak m\to N_\mathfrak m$ were injective for each $\mathfrak m$, then the kernals $\ker\psi_\mathfrak m$ would be zero as well. But then $0=\ker\psi_\mathfrak m = (\ker\psi)_\mathfrak m$ for each $\mathfrak m$ since localization is exact, hence $\ker\psi=0.$ By considering the cokernal we can also conclude the same about surjectivity. Hence $\psi_\mathfrak m$ is an iso for each $\mathfrak m$ implies $\psi$ is an iso.
