For $V$ vector space, $W$ subspace, $V \simeq W \oplus V/W$?

Question

I thought this seemed intuitively true. So I searched for it but never found it. I thought that if it were true, surely it would appear somewhere in the Rank-Nullity Theorem page on Wikipedia. But no luck. So I tried to prove it and I think I have done. But I'm obviously very sceptical about my own work here. Is this correct ?

Let $V$ be a vector space over some field $K$ and let $W$ be a subspace of $V$.
Then $V/W$ is a vector space over $K$, so it has a basis $\mathcal B$. Define a linear map $ \phi : V/W \rightarrow V$ by choosing $\phi(x)$ for any $x \in \mathcal B$. More precisely, do it such that $\phi(x) \in x \quad \forall x \in \mathcal B$, in which case this property extends to all $x \in V/W$

For variables in $V$ I shall use $v$ instead of $x$ to avoid confusion. Let $p$ be the projection onto $V/W$.
My claim is that $f : \begin{array}{cccc} V & \rightarrow & W \oplus V/W \\ v & \mapsto & (v-\phi \circ p (v),\, p(v)) \end{array}$ is an isomorphism.

As an ordered pair of sums and compositions of linear maps, $f$ is linear.

Let's prove that $f$ is injective. $f(v) = 0 \iff (\,p(v)=0 \,\text{and}\, v- \phi \circ p(v)=0\,) \implies v-\phi(0)=0 \implies v=0$

$f$ is onto since for any $(\alpha,\, y) \in W \oplus V/W$ we have $f(\alpha + \phi(y))=(\alpha+\phi(y)-\phi \circ p (\alpha + \phi(y)),\, p (\alpha + \phi(y))\,)$

On the right hand side of the ordered pair, $p(\alpha)=0$ since $\alpha \in W$, and also $p(\phi(y))=y$ as $\phi(y)\in y \in V/W$ so in total we're left with $y$.
As for the left hand side, we $\alpha + \phi(y) - \phi(\text{the right hand side})$ which adds up to $\alpha$, QED.

Answer 1

This is true, irrespective of $V$ being finitely generated or not, but it relies on the axiom of choice, that's used to prove every subspace has a complement.

If you accept the existence of a complement (which usually people do), then you have a projection along it and you get the result. Namely, if $W'$ is a subspace so that $W\cap W'=\{0\}$ and $W+W'=V$, you can define $p\colon V\to W$ by $p(v)\in W$ and $v-p(v)\in W'$. Observe $p$ is linear.

Then you can define $g\colon V\to W'$ with $g(v)=v-p(v)$; this map is surjective and has kernel $W$, so it induces an isomorphism $q\colon V/W\to W'$.

Then define $f\colon V\to W\oplus V/W$ by $$ f(v)=\bigl(p(v),q^{-1}(v-p(v))\bigr) $$ and prove it is an isomorphism.

Answer 2

Vector spaces are isomorphic iff they have the same dimension.

Answer 3

Let $V$ be a vector space over field $K$ and $W$ be a $K$-sub space of $V$.

Now let us consider the natural surjective map $\eta : V \rightarrow V/W$, where $\eta (x) = x + W$.

Note that, ker $\eta = W$.

Let $\{ x_{i} + W \}_{i \in I}$ be a $K$-basis of $V/W$.

Now let $y_{i} \in V$ such that $\eta(y_{i}) = x_{i} +W$ for all $i \in I$.

Note that $y_{i}$'s are linearly independent and let $W'$ be the subspace spanned by $y_{i}$'s.

Extend the set of linearly independent vectors $A = \{ y_{i} \}_{i \in I}$ to a basis set $B$ of $V$.

Note that $B \setminus A$ is a basis set of $W$.

So, $V = W \oplus W'$.

And $W' \cong V/W$ implies that $V \cong W \oplus V/W$.