# For modules, $\text{Hom}$ is an exact functor

To paraphrase Serge Lang, Algebra: Proposition 2.1.

Let $$X$$, $$X^\prime$$, $$X^{\prime\prime}$$ and $$Y$$ be modules over some ring $$A$$. (The ring does not change, and will therefore not be mentioned anymore to avoid notational overload) The sequence $$X^\prime \xrightarrow{\lambda} X \xrightarrow{\mu} X^{\prime\prime} \rightarrow 0 \tag{1}$$ is exact if and only if the sequence $$0 \rightarrow \text{Hom}(X^{\prime\prime}, Y) \xrightarrow{m_\mu} \text{Hom}(X, Y) \xrightarrow{m_\lambda} \text{Hom}(X^\prime, Y) \tag{2}$$ is exact for all $$Y$$. (Lang uses the notation $$\text{Hom}(\lambda, Y)$$ to denote what I call $$m_\lambda$$, but I think the latter is more notationally convenient)

Proof.

Suppose the first sequence is exact. Consider a homomorphism $$g: X \rightarrow Y$$ such that $$X^\prime \xrightarrow{\lambda} X \xrightarrow{g} Y$$ is $$0$$. Then $$g$$ must vanish on the image of $$\lambda$$, implying that we can factor it through $$X/\text{Im}$$ $$\lambda$$. But since $$X\rightarrow X^{\prime\prime}$$ is surjective, we have an isomorphism $$X/\text{Im}$$ $$\lambda\cong X^{\prime\prime}$$, so that we can factor $$g$$ through $$X^{\prime\prime}$$, which shows that the kernel of $$m_\lambda$$ is contained within the image of $$m_\mu$$. The rest of the proof is then left to the reader.

From my understanding of exact sequences, what is now left to prove is that the image of $$m_\mu$$ is contained within the kernel of $$m_\lambda$$, and that $$m_\mu$$ is injective, as well as the entire converse statement. We begin by explicitly writing the action of $$m_\mu$$ (resp. $$m_\lambda$$) on an element $$h\in\text{Hom}(X^{\prime\prime},Y)$$: $$m_\mu(h) = h\circ\mu \in \text{Hom}(X,Y)$$ First we show that $$m_\mu$$ is injective. Let $$k = m_\mu(h)$$ and $$k^\prime = m_\mu(h^\prime)$$ be elements of $$\text{Hom}(X,Y)$$. Let $$k = k^\prime$$. We then have: $$h-h^\prime = k\circ\mu - k^\prime\circ\mu = (k-k^\prime)\circ\mu = 0\circ\mu = 0 \tag{3}$$ showing that if $$k = k^\prime$$, then $$h = h^\prime$$ and $$m_\mu$$ is surjective. The last equality in $$(3)$$ follows from the fact that $$\mu$$ is surjective.

Now we show that $$\text{Im}(m_\mu) \subset \text{ker}(m_\lambda)\Leftrightarrow m_\lambda(m_\mu(h)) = 0$$ for all $$h \in \text{Hom}(X^{\prime\prime},Y)$$. We know that since $$(1)$$ is exact, $$\mu(\lambda(x)) = 0$$ for all $$x \in X^\prime$$. Let $$h\in\text{Hom}(X^{\prime\prime}, Y)$$ and $$f = h \circ \mu \circ \lambda\in\text{Hom}(X^\prime,Y)$$. But then $$f(x) = h(\mu(\lambda(x)) = h(0)$$, which must be $$0$$ since $$h$$ is a module-homomorphism. But since this must hold for all $$x\in X^\prime$$, we have $$h\circ\mu\circ\lambda\equiv m_\lambda(m_\mu(h)) = 0$$ and therefore the image of $$m_\mu$$ is contained in the kernel of $$m_\lambda$$.

However, I struggle really hard with the converse implication. For starters, I would like to know if I can conclude the surjectivity of $$\mu$$ from the injectivity of $$m_\mu$$, by inverting the reasoning in $$(3)$$ (essentially putting the zero on the left side of the expression and concluding that if $$0\circ\mu = 0$$, then $$\mu$$ must be surjective). For the equality of $$\text{Im}$$ $$\lambda$$ and $$\text{ker}$$ $$\mu$$, some pointers about which direction to go for a proof would be fine, I do not need/want a complete walkthrough.

• Lang’s Prop 2.1 says that the sequence (2) should be exact for all $Y$. That makes a big difference. Commented Apr 7 at 11:41
• that's right, i forgot to write that in my question. i will edit Commented Apr 7 at 11:43

Hom is usually not an exact functor. That may just be a typo in your title but it's extremely important to get that right. $$\hom(A,-)$$ is right exact and so is $$\hom(-,B)$$ if you value it in $$\mathsf{RMod}$$ ($$\hom(-,B):\mathsf{RMod}^{\mathsf{op}}\to\mathsf{RMod}$$).
What does it mean for $$m_\mu$$ to inject? It means a morphism $$f:X''\to Y$$ is zero if $$f\mu=0$$; as commented, we need to assume this for all $$Y$$ and now it is well known that means $$\mu$$ must be surjective. For example let $$f$$ be the quotient $$X''\to X''/\mathrm{Im}(\mu)$$. This idea is actually very fruitful and prompts the solution directly.
What does it mean for $$\ker m_\lambda=\mathrm{im}\,m_\mu$$ for all $$Y$$? To let that run for all $$Y$$ it is clear $$\mu\lambda=0$$ holds (plug in $$Y=X'$$). It tells as $$f\lambda=0$$ iff. $$f=g\mu$$, where $$f:X\to Y$$ is some map and $$g:X''\to Y$$ is some map. We would take $$f:X\to X/\mathrm{Im}(\lambda)$$ the quotient, using the same idea, and $$f\lambda=0$$ of course so we know $$f$$ factors as $$X\to X''\to X/\mathrm{Im}(\lambda)$$. We know $$\mu$$ surjects and $$\mu\lambda=0$$ so it follows there is a section $$X/\mathrm{Im}(\lambda)\to X''$$ coming from $$\mu$$ which must both inject and surject (as we know it has some mysterious right inverse $$g$$) i.e. is an isomorphism. Therefore the kernel of $$\mu$$ is exactly the kernel of $$X\to X/\mathrm{Im}(\lambda)$$, which is $$\mathrm{Im}(\lambda)$$ - as desired.
By the way, the property for a $$B$$ so that $$\hom(-,B)$$ of a sequence is short exact iff. the original is short exact, is that $$B$$ should be faithfully injective. The property that $$\hom(-,B)$$ of a short exact sequence is short exact is that $$B$$ is injective. Most modules are not injective; $$\hom$$ is not exact.