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

Question

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.

Answer 1

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.