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I am trying to prove the following proposition rigorously given at Michael Atiyah commutative algebra book. There is one step which I am currently stuck in. Suppose that $N,N^{\prime}, \ and \ N^{\prime \prime}$ are A-modules.

The following exact sequence $$0 \xrightarrow \ N^{\prime} \xrightarrow{u} N \xrightarrow{v} N^{\prime \prime}$$ is exact iff for all A-modules M we have that the following sequence is exact $$0 \xrightarrow \ Hom(M,N^{\prime}) \xrightarrow{\bar{u}} Hom(M,N) \xrightarrow{\bar{v}} Hom(M,N^{\prime \prime}).$$

I proved the reverse direction, however in the forward direction I have one step I am currently stuck in. Here is what I have so far.

So we want to show that $ker(\bar{v}) = im(\bar{u})$. I showed the inclusion $im(\bar{u}) \subset ker(\bar{v})$. Suppose that $\psi \in ker(\bar{v})$. Then, we have $\bar{v}(\psi) = v \circ \psi = 0$. Then $im(\psi) \subset ker(v) = im(u) \cong N^{\prime}$. Then, we can reinterpret $\psi : M \rightarrow N$ as a morphism $\phi : M \rightarrow N^{\prime}$. Denote this new map by $\phi$. Now I am having troubles showing that $\bar{u}(\phi) = \psi$. Please maybe provide a hint as I would like to discover this by myself.

In my question I am not asking for a proof. I am asking to check certain aspect of my argument and improving on that. Thus, this post isn't a duplicate.

gymbvghjkgkjkhgfkl
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Hint: The isomorphism $\mathrm{im}(u)\cong N'$ is given by $u^{-1}$, so $\phi=u^{-1}\circ\psi$.

($u^{-1}$ is restricted to having domain $\mathrm{im}(u)$, but that's no problem since $\mathrm{im}(\psi)\subseteq\mathrm{im}(u)$.)

anon
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