How is $g(\mu(m))= f(m)$ when $\mu(m)=n-\rho(\nu(n))$?
2026-03-29 11:44:10.1774784650
Is there an error in this proof showing dual functor over free module is exact?
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We have that $\mu(m) - \rho(\nu(\mu(m))) = \mu(m) - \rho(0) = \mu(m)$ by exactness of the first sequence $0 \rightarrow M \rightarrow N \rightarrow P \rightarrow 0$.
Edit: To show that $g(\rho(\nu(n)) = 0$, note that $g(n) = f(m)$, with $m$ the unique element of $M$ such that $\mu(m) = n - \rho(\nu(n))$. So $\rho(\nu(n)) - \rho(\nu(n)) = 0 = \mu(0)$. Hence $g(\rho(\nu(n)) = f(0) = 0$.