# Exact sequence of $Hom(, M)$.

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 $$\phi = \eta\beta$$?

• I don't understand what you mean by "setting $M$ to be the modulo of $\operatorname{im}(\alpha)$". Commented May 6, 2018 at 8:04
• @Ennar, I meant $M = B/im(\alpha)$. Commented May 6, 2018 at 11:44
• That doesn't make sense, the statement is true for any $M$, not just a particular one. What is true is that $C\cong B/\operatorname{im}\alpha$, which is just the first isomorphism theorem. Commented May 6, 2018 at 11:58

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, $s = 0$.

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

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

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

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.

• 1. But I thought $s\beta = 0 \implies \beta = 0$ only if $\beta$ is injective like you said for $f$. Commented May 6, 2018 at 11:41
• @Hawk, no, you've got the definition wrong. Remember the definition for injective function $f(x) = f(y)\implies x = y$. Well, it's the same definition in any category when you replace elements $x$ and $y$ for morphisms $x$ and $y$. In Abelian category, this simplifies to $fx = 0 \implies x = 0$ which literally says that $f$ is mono if and only if its kernel is trivial. This should be familiar characterization from linear algebra, group/ring theory, etc. Epimorphisms are defined dually, $f$ is epi iff $xf = yf\implies x = y$ and in Abelian category it simplifies to $xf = 0\implies x = 0$. Commented May 6, 2018 at 11:49
• Oh, I noticed a mistake which might have confused you. I wrote: "$gf = 0\implies g = 0$ if and only if $f$ is monomorphism" where I should have written "$gf = 0\implies g = 0$ if and only if $f$ is epimorphism." I've corrected the mistake, sorry about that, @Hawk. Commented May 6, 2018 at 11:53
• There are still things not clear to me. I don't know all the jargons in category theory 1. If $g = 0$, then $0f = 0$ no matter what $f$ is, so why does that imply $f$ is surjective...? 2. Also you wrote "$\beta$ is the cokernel of $\alpha$", you meant to write $C = B/im(\alpha)$ right? Because cokernels are sets, not maps. Commented May 6, 2018 at 22:19
• imgur.com/a/Ehk9zTL this is so far what I am getting, I am not seeing how this induces the dotted map. Commented May 6, 2018 at 23:31

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 $\psi\beta = \phi$.

• So in the final step, we set $M = B/Im(\alpha)$? This is what I was wondering, why are we allowed to choose $M$ to be the cokernel? Commented May 6, 2018 at 11:41
• Are we just concerned with the existence of a single $\eta$ from ever possible choice of $M$? Commented May 6, 2018 at 11:46
• So this is what I am understanding imgur.com/a/Ehk9zTL . I don't see how $\psi$ is induced from this. Commented May 6, 2018 at 22:41
• @Hawk No, I don't set $M = B/im(\alpha)$... I say that since $\phi\alpha = 0$, then $\phi$ vanishes on $im(\alpha)$, therefore $\phi$ factors through the quotient $B/im(\alpha)$. Commented May 7, 2018 at 10:01