I have few questions, some trivial probably, but I am a beginner in homological algebra, so I'd be greatful for any explanations:

1. If $D$ is a rational vector space, $F$ is the base and $G$ is an abelian group such that $G\approx F/R $ for some $R$ (we can consider $F$ as a free group on the elements of $G$). Then $G\approx F/R\subset D/R$, why $D/R$ is divisible?

2. Why from the fact that quotient of the divisible group is divisible (and hence injective), we know that for any abelian group $G$ there is an exact sequence: $$0\rightarrow G\rightarrow I\rightarrow J\rightarrow 0,$$ for $I,J$ injective.

3. If $P$ is projective and $0\rightarrow G\rightarrow I\rightarrow J\rightarrow 0$ is an injective resolution, then $Hom(P,I)\rightarrow Hom(P,J)$ is onto. Why?


2 Answers 2


1) $D/R$ is divisible because $D$ is a rational vector space, i.e. $D\simeq \bigoplus\mathbb{Q}$ as an abelian group. And this group is divisible. And quotients of divisible groups are divisible.

2) That's because every abelian group can be embedded into a divisible group. That's pretty much what you've shown in 1. You just have to start from the other end, i.e. you take an abelian group $G$, take its presentation $G\simeq F/R$ for some free abelian group $F=\bigoplus\mathbb{Z}$ and embed $F$ into $D=\bigoplus\mathbb{Q}$.

So take an embedding $f:G\rightarrow I$ into a divisible group and define $J:=I/\mbox{im}(f)$. The natural quotient map yields the short exact sequence you are looking for:

$$0\rightarrow G\rightarrow I\rightarrow J\rightarrow 0$$

Note that $J$ is injective because it is divisble (as a quotient of divisible group).

3) This doesn't seem to be dependent on injectivness at all. If $P$ is projective then $\mbox{Hom}(P, -)$ is an exact functor, so it takes exact sequences into exact sequences, so we have an exact sequence

$$0\rightarrow \mbox{Hom}(P, G)\rightarrow \mbox{Hom}(P, I)\rightarrow \mbox{Hom}(P, J)\rightarrow 0$$

That's why the induced map $\mbox{Hom}(P, I)\to\mbox{Hom}(P, J)$ is onto.


Question 2: Any abelian group $G$ can be embedded into an injective $\mathbb{Z}$-module $I$ using the dual module ${\rm Hom}_{\mathbb{Z}} (M,\mathbb{Q}/\mathbb{Z})$. Hence we obtain a short exact sequence $$ 0\rightarrow G \rightarrow I \rightarrow I/G \rightarrow 0. $$ Since $I$ is injective, also the quotient $J=I/G$ is injective. Since we are talking about $\mathbb{Z}$-modules, divisibility is the equivalent to injectivity. Therefore the "fact that quotient of the divisible group is divisible" works.


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