When does $V/\operatorname{ker}(\phi)\simeq\phi(V)$ imply $V\simeq\operatorname{ker}(\phi)\oplus\phi(V)$?

Solution 1:

It's not true in general. (In fact, it's false in most of the common categories: groups, rings, modules, etc; it just happens to hold for finite-dimensional vector spaces because dimension completely classifies them.) To take two examples:

• Groups: The symmetric group $S_3$ contains a normal subgroup $A_3 = \mathbb{Z}_3$, which is the kernel of the sign map $f:S_3 \to \mathbb{Z}_2$. But $S_3$ is not the direct product of $A_3$ and some other $N\subset S_3$; the group $N$ must then have order $2$ and thus be isomorphic to $\mathbb{Z_2}$, but clearly $S_3\not = \mathbb{Z}_3 \oplus\mathbb{Z}_2$.
• Modules: The quotient map $f:\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ has kernel $n\mathbb{Z}$, but $n\mathbb{Z}$ is not a direct summand of $\mathbb{Z}$. For if $\mathbb{Z} = n\mathbb{Z} \oplus M$, then $M =\mathbb{Z}/n\mathbb{Z}$, but $\mathbb{Z}$ is torsion-free.

Solution 2:

$V\simeq ker(\phi)\oplus \phi(V)$ is always true for a linear map $\phi:V\rightarrow W$, even if $V$ and $W$ are infinite dimensional.

Prop: Let $\phi:V\rightarrow W$ be a linear map. There exists a subspace $R$ of $V$ such that

1. $R\oplus\ker(\phi)=V$
2. $\phi:R\rightarrow\phi(V)$ is an isomorphism.

Proof: Choose a basis $\{w_i,\ i\in I\}$ for the subspace $\phi(V)$. Let $v_i\in V$ be such that $\phi(v_i)=w_i$.

Notice that $\{v_i,\ i\in I\}$ is a linear independent set. Let $R$ be the span of $\{v_i,\ i\in I\}$.

Consider $\phi:R\rightarrow\phi(V)$. It is obvious that $\phi$ is surjective.

Now, suppose that $r\in R\cap\ker(\phi)$. Since $r=\sum_{k=1}^m a_kv_{i_k}$ then $0=\phi(r)=\phi(\sum_{k=1}^m a_kv_{i_k})=\sum_{k=1}^m a_kw_{i_k}$. Thus $a_{i_1}=\ldots=a_{i_m}=0$ and $r=0$. Therefore $R\cap\ker(\phi)=\{0\}$ and $\phi:R\rightarrow\phi(V)$ is injective. Thus $\phi:R\rightarrow\phi(V)$ is an isomorphism.

Now, let $v\in V$. Let $\phi(v)=\sum_{s=1}^n b_sw_{i_s}$. Consider $v-\sum_{s=1}^n b_sv_{i_s}$. Notice that $\phi(v-\sum_{s=1}^n b_sv_{i_s})=0$. Notice that $\sum_{s=1}^n b_sv_{i_s}\in R$. So every $v\in V$ is a sum of a vector in $R$ and a vector in $\ker(\phi)$.

Solution 3:

Another way of asking this more generally: When, for any $A,B$, do there exist no nontrivial extensions $0\to A\to E\to B\to 0$?

From homological algebra (at least in the case of an abelian category, perhaps more generally), we can say that this is true when every object in our category is projective.

The relevant fact about vector spaces (modules over a field) is that they all have a basis. So they are all free, hence projective.

The same holds if we consider the category of $R$-modules, where $R$ is a (not necessarily commutative) semisimple artinian ring.

Solution 4:

The reason the map $\log$ has a (noncanonical) splitting is the following:

An algebraic number all of whose Galois conjugates have absolute value $1$ is a root of unity.

It follows immediately from this that the kernel of $\log: R^\times \to \mathbf R^{r+s-1}$ consists precisely of $\text{Tors}(R^\times)$. Moreover, $\log$ is an isomorphism of $R^\times/\text{Tors}(R^\times)$ with a lattice of $\mathbf R^{r+s-1}$. Therefore $R^\times/\text{Tors}(R^\times)$ is finitely generated, and since $\text{Tors}(R^\times)$ is finite, so is $R^\times$; it follows from the structure theorem for finitely generated groups that $R^\times \cong \text{Tors}(R^\times) \oplus R^\times/\text{Tors}(R^\times)$ (noncanonically).