# existence of an exact complex

Let $R$ be a commutative ring and let $L, M, N$ be three $R-$modules. Given an exact sequence of $R-$modules $$0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0.$$ Show that if $T$ is a free $R-$module, then one obtains an exact complex $$0 \rightarrow \textrm{Hom}_{R}(T,L) \xrightarrow{\alpha \circ} \textrm{Hom}_{R}(T,M) \xrightarrow{\beta \circ} \textrm{Hom}_{R}(T,N) \rightarrow 0.$$

Edit: How to explain for every $R-$module you obtain the same result except the $\rightarrow 0$ part ? While studying for my preliminaries I came across this question. Any help in solving this is much appreciated.

• @MarianoSuárez-Álvarez, thank you for the suggestion. I haven't tried that. This was bugging me all day. – user358174 Nov 23 '16 at 3:13
• Left-exactness of the complex holds regardless of whether $T$ is a free module. The special fact we need is that every element of $\operatorname{Hom}_R(T,N)$ "lifts" to an element of $\operatorname{Hom}_{R}(T,N)$ so that $\beta \circ$ is surjective. That is, what's special here is that $T$ is projective. – Omnomnomnom Nov 23 '16 at 3:13

Note that $\textrm{Hom}_{R}(T,-)$ is a covariant functor and left exact functor. In particular, suppose that $T$ is a free $R$-module. Then $T$ is also a projective module, and then $\textrm{Hom}_{R}(T,-)$ is a covariant functor and exact functor, since $\textrm{Hom}_{R}(T,-)$ maps surjective morphisms to surjective morphisms.

• How would you go about doing this for every $R$-module $T$, rather than just when $T$ is a free $R$-module? – user389056 Nov 23 '16 at 3:27
• For any $R$-module $T$, we only obtain $$0 \to \textrm{Hom}_R(T,L) \overset{\alpha_*}{\to} \textrm{Hom}_R(T,M) \overset{\beta_*}{\to} \textrm{Hom}_R(T,N)$$. – bing Nov 23 '16 at 3:33
• @bing, thank you very much. This really helps. – user358174 Nov 23 '16 at 7:46
• @ ManMath It is my pleasure. – bing Nov 23 '16 at 7:52
• For $\alpha: L \to M$, $\alpha_*=\textrm{Hom}_{R}(T,\alpha): \textrm{Hom}_{R}(T,L) \to \textrm{Hom}_{R}(T,M)$ sends $f:T\to L$ to $\alpha \circ f$, which implies that $\alpha_*$ is injective. Then $$\beta_* \circ \alpha_* (f)= \beta \circ \alpha \circ f = 0 \Longrightarrow \textrm{Im} \alpha_* \subseteq \textrm{ker} \beta_*.$$ For any $g \in \textrm{ker} \beta_*$, we have $\beta_*(g) = \beta \circ g =0$, so $g = \alpha \circ u = \alpha_*(u)$ by using the universal property of the kernel of $\beta$. Hence $\textrm{Im} \alpha_* = \textrm{ker} \beta_*$. – bing Nov 25 '16 at 1:59