Exact sequence of $Hom(, M)$.

Question

Yes, I know there is a duplicate somewhere, but I don't quite follow their solutions. Here is the problem:

$$A \stackrel{\alpha}{\to} B \stackrel{\beta}{\to} C \to 0$$ is exact and I want to show that

$$0\to \text{Hom}(C,M) \stackrel{F}{\to} \text{Hom}(B,M) \stackrel{G}{\to} \text{Hom}(A,M)$$ is exact,

with $F(\eta) = \eta \beta$ and $G(\phi) = \phi\alpha$. I don't know how to draw commutative diagrams.

(i) $\ker(F) = \{ F(\eta) = 0\} = \{ \eta \beta = 0 \}$. Now $\beta$ is onto, so $\eta c = 0$ for all $c \in C$, so $\eta = 0$.

(ii) Suppose $\phi \in \operatorname{Im}(F)$, then $\exists \beta \in \operatorname{Hom}(C,M)$ such that $\phi = \eta \beta$. Then $G(\phi) = G(\eta \beta) = \eta \beta \alpha = \eta 0 = 0$ because exactness tells us $\alpha \in \ker \beta$. So $\operatorname{Im} \subset \ker G$.

Here is the part that is tripping almost everyone: the other inclusion $\ker G \subset \operatorname{Im}F$.

All of the answers I've read so far involve setting $M$ to be the modulo of $\operatorname{Im}(\alpha)$. Why are we allowed to change $M$?

Also, I wrote something else that I don't know if it works.

I say consider $\phi\in \ker G$, then for the map $\eta\beta - \phi$ applied to $\alpha$, we get $(\eta\beta - \phi)\alpha = \eta\beta\alpha - \phi\alpha = \eta\beta\alpha - 0 = 0 - 0$. The first $0$ comes from definition and the second comes from repeated argument from above from exactness that $\operatorname{Im}\alpha = \ker\beta$. So doesn't this show $\phi = \eta\beta$?

Answer 1

Let $\phi \in \ker(G)$, i.e. $\phi : B \to M$ is such that $\phi\alpha = 0$. We want to find $\psi : C \to M$ such that $\psi\beta = \phi$.

The sequence $A \to B \to C \to 0$ is exact, therefore $\beta : B \to C$ is surjective and its kernel is $\operatorname{im}(\alpha)$. It's an easy exercise in linear algebra to check that this implies that $\beta$ induces an isomorphism $B/\operatorname{im}(\alpha) \cong C$. I suppose that this is what's bothering you?

Then the fact that $\phi\alpha = 0$ means that $\phi$ vanishes on $\operatorname{im}(\alpha)$. It follows that $\phi$ factors through the quotient $B/\operatorname{im}(\alpha)$, which is $C$, hence there exists $\psi : C \to M$ such that $\psi\beta = \phi$.

Answer 2

First of all, as I've said in the comments, I don't understand what you mean by "setting $M$ to be the modulo of $\operatorname{im}\alpha$", so I can't help you there.

Regarding your last paragraph, it doesn't work. In Abelian category $gf = 0\implies g = 0$ if and only if $f$ is epimorphism. For all we know, your $\alpha$ could be zero map.

Now, to the proof.

1. $F$ is mono:

Let $Fs = 0$. Then $s\circ \beta = 0$ and since $\beta$ is epimorphism, $s = 0$.

2. $G\circ F = 0$ (equivalently, $\operatorname{im}F\subseteq \ker G$):

$(G\circ F)(s) = s\circ \beta \circ \alpha = 0$ since $\beta\circ\alpha = 0$.

3. $ \ker G\subseteq\operatorname{im}F$:

Let $Gs = 0$, i.e. $s\circ\alpha = 0$. Note that $\beta$ is cokernel of $\alpha$, so by the universal property of cokernel, there exists $s'$ such that $s'\circ\beta= s$. Thus $s\in\operatorname{im}F$.

EDIT: Let me elaborate on the last paragraph. Let's say we work with modules or Abelian groups, it works the same in any Abelian category (just without elements).

Let's start by noting that $B/\ker\beta\cong \operatorname{im}\beta = C$ by the first isomorphism theorem. Write $\varphi\colon C\to B/\ker\beta$ for the inverse of the map $b+\ker\beta\mapsto \beta(b)$. Immediately we have that the composition $\varphi\circ \beta$ is the canonical epimorphism $B \twoheadrightarrow B/\ker\beta .$ Now, since $\ker\beta=\operatorname{im}\alpha$ (it's strict equality, not just isomorphism), we can write $B/\operatorname{im}\alpha$ instead of $B/\ker\beta$.

Now, let $s\colon B\to M$ be such that $s\circ\alpha = 0$. I claim that there exists unique map $t\colon B/\operatorname{im}\alpha\to M$ to make the following diagram commute: $\require{AMScd}$ \begin{CD} A @>\alpha>> B @>\beta>> C @>>> 0\\ @| @| @V\varphi VV \\ A @>\alpha>> B @>\small\text{canon. epi}>> B/\operatorname{im}\alpha @>>> 0\\ @. @VsVV @VtVV \\ @. M @= M \end{CD}

Define $t(b+\operatorname{im}\alpha) = s(b)$. I will leave the verification that this is well-defined to you. Uniqueness of such a map is obvious from the definition.

Finally, define $s'\colon C\to M$ by setting $s' = t\circ \varphi$.

All in all, what we have just proved is the following:

Let $A\stackrel{\alpha}{\to} B \stackrel{\beta}{\to} C \to 0$ be an exact sequence. For every $s\colon B\to M$ such that $s\circ\alpha = 0$ there exists unique $s'\colon C\to M$ such that $s'\circ\beta = s$.

This is called the universal property of cokernel. In this case $\beta$ is the cokernel of $\alpha$. More precisely, the pair $(C,\beta)$ is the cokernel, but we often omit mentioning the object since it is understood from the context.

Thus, cokernel is not just an object, it is a pair of an object and a map onto it. Canonical choice for cokernel is $B \twoheadrightarrow B/\operatorname{im}\alpha $ where the arrow is the canonical epimorphism. But, in our example, we used that this canonical cokernel is isomorphic to $C$ since we really wanted a map from $C$, not from $B/\operatorname{im}\alpha$.

But, it is not enough that $B/\operatorname{im}\alpha \cong C$ to say that $C$ is cokernel, the maps onto $C$ and $B/\operatorname{im}\alpha$ must commute with the isomorphism, as you can see from the diagram. The point is, objects themselves don't matter much on their own, it's the maps that do.