I want to

show that if for all modules $D$ the sequence $$0 \to \operatorname{Hom}_{R}(N,D) \xrightarrow{\varphi^*} \operatorname{Hom}_{R}(M,D) \xrightarrow{\psi^*} \operatorname{Hom}_{R}(L,D) \to 0$$ is exact, then the sequence $$0 \to L \xrightarrow{\psi} M \xrightarrow{\varphi} N \to 0$$ is split exact.

Now, we already know that, for all $R$-modules $D$, $$0 \to \operatorname{Hom}_{R}(N,D) \xrightarrow{\varphi^*} \operatorname{Hom}_{R}(M,D) \xrightarrow{\psi^*} \operatorname{Hom}_{R}(L,D)$$ is exact if and only if the sequence $$L \xrightarrow{\psi} M \xrightarrow{\varphi} N \to 0$$ is exact.

Now, I think that the sequence $$0 \to L \xrightarrow{\psi} M \xrightarrow{\varphi} N \to 0$$ is split exact is equivalent to the condition that there exists a retraction $$r:M \to L$$ such that $$r \circ \psi = \operatorname{Id}_{L}$$.

We pick $$D = L$$ since by assumption this should hold for all R-modules $D$, so in particular $L$. Now, by assumption, $\psi^{*}$ is surjective, hence for all $\ell \in \operatorname{Hom}_{R}(L,L)$ there exists an $f \in \operatorname{Hom}_{R}(M,L)$ such that $$\psi^{*}(f) = f \circ \psi = \ell.$$ Now, just pick the identity homomorphism $$\operatorname{Id}_{L} \in \operatorname{Hom}_{R}(L,L).$$ By the surjectivity of $\psi^{*}$, there exists an $$f' \in \operatorname{Hom}_{R}(M,L)$$ such that $$\psi^{*}(f') = f' \circ \psi = \operatorname{Id}_{L}$$ i.e. $f'$ is the retraction we were looking for.

Is this correct? I think I may be missing something more I must show, I am not sure, although I think what I´ve shown is that $\psi$ is injective (since a retraction is atleast in some categories, the same as a left-inverse) and this together with the equivalence between retractions and split exactness, together with what I wrote earlier should show (unless I am mistaken) that the sequence in question is split exact.

Comment: $\psi^{*}$ denotes what I believe in category-theory would be called a pullback, i.e. precomposition on each element in its domain/source (here $R$-module homomorphisms from $M$ to $L$).

Edit: I think one problem is that the "split-exactness iff exists retraction" statement can only be applied on a SES, here I don´t know that $$0 \to L \to M \to N \to 0$$ is a short exact sequence.

  • $\begingroup$ Thanks for the feedback Mariano :) You mean I should shorten it? Perhaps, although I think I recently got some feedback from a moderator to provide more context in a question I asked, so it is quite hard to strike the right balance. But point taken. $\endgroup$
    – Ben123
    Commented Sep 30, 2023 at 1:40
  • $\begingroup$ I have edited my question, I believe I missed the statement about my choice of setting $D = L$, perhaps this is what you meant by "You are probably missing a quantifier on $D$"?. $\endgroup$
    – Ben123
    Commented Sep 30, 2023 at 1:43
  • $\begingroup$ Oh I see, you meant in the beginning. Someone fixed it. Anyways :) $\endgroup$
    – Ben123
    Commented Sep 30, 2023 at 1:44


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