Extension of vector spaces and abelian groups

Question

While reading about modules from Hilton & Stammbach's Homological algebra, I saw the following statement : $\Lambda$ is a ring.

$\Lambda$ modules are generalizations of vector spaces and abelian groups. If $\Lambda$ is a field then $\Lambda$ module is a vector space and if $\Lambda=\mathbb{Z}$ then $\Lambda$ module is an abelian group.

I understand this quite well. Here comes the next statement.

In theory of vector spaces there is no interest in the following question : Given vector spaces $A$ and $B$ over a field $K$, find all vector spaces $E$ over $K$ such that $E/B\cong A$ as any such $E$ is isomorphic to $A\oplus B$. However this question turns quite interesting in case of abelian groups $A,B,E$.

First of all i do not understand how does $E\cong A\oplus B$.. I can think of one map $E\rightarrow E/B\rightarrow A\rightarrow A\oplus B$ first map being the quotient map, second map being the isomorphism and third map being inclusion.. How ever i do not think this is a surjective map.

So, what is the isomorphism that they are talking about $E\cong A\oplus B$..

What is stopping them to do the same for abelian groups..

Answer 1

There isn't a unique isomorphism; if you're given a short exact sequence $0 \to A \to E \to B \to 0$ of vector spaces, to exhibit an isomorphism $E \cong A \oplus B$ from here you need to pick a splitting, which means a section $s : B \to E$ of the map $E \to B$. Given a splitting, the isomorphism $A \oplus B \to E$ has components the given map $A \to E$ and the splitting.

Splittings need not exist for general modules, and in particular don't exist for abelian groups in general: the smallest example is

$$0 \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_2 \to 0.$$

Splittings always exist for vector spaces (assuming the axiom of choice) because vector spaces have bases, and so are free.

Answer 2

If $E, B, A$ are vector spaces such that $E/B \cong A$, this means we have a surjective map $\pi \colon E \rightarrow A$ (the quotient map) whose kernel is precisely $B$. Now, in the category of vector spaces, given a vector space $E$ and a subspace $B \leq E$, we can always find another subspace $B' \leq E$ such that $B \oplus B' = E$ (this is an internal direct sum). To do that, we can (assuming the axiom of choice, if needed) choose a basis for $B$, extend it to a basis of $E$ and let $B'$ be the vector space spanned by the basis elements that do not belong to $B$.

Having found such a $B'$, we can see that $\pi|_{B'} \colon B' \rightarrow A$ is an isomorphism and so

$$E = B \oplus B' \cong B \oplus A = \ker(\pi) \oplus \mathrm{im}(\pi). $$