In the textbook I am reading (A Course in Modern Mathematical Physics by Szekeres), the definition of the quotient space of a vector space $V$ given a subspace $W$, $V/W$, is motivated by the desire to find a way to construct complementary subspaces to that fixed subspace $W$ (ie. it is defined after giving the typical nonconstructive proof of the existence of complementary subspaces using Zorn's Lemma).
The result about all complementary subspaces to $W$ being isomorphic to $V/W$ and thus to each other is proved, and then it is stated that "the quotient space is a method for constructing the 'canonical complement' to the subspace".
Now it's quite clear to me how one can construct the quotient space from a complementary subspace using the explicit isomorphism (called the quotient map by Axler, $T: U \to V/W$ defined by $T(u) = u +W$), but the converse is not clear. Obviously things are not that simple since a given coset is just a set and is not given to us in the form $u+W$ for $u \in U$, with $U$ the complementary subspace which we want to construct! There is no clear $T^{-1}$ to use.
Therefore, my question is, given a quotient space, how does one in general construct a complementary subspace? Answers which give an example of such a construction and/or mention applications in which such a construction is useful would be very much appreciated!
Best Answer
Question: "Therefore, my question is, given a quotient space, how does one in general construct a complementary subspace?"
Answer: If $k$ is a field and $V$ is a $k$-vector space, it follows (if you accept the "axiom of choice") that $V$ has a basis $\{v_i\}_{i\in I}$. From this property it follows that $V$ is a projective $k$-module: For any surjective map
$$\rho: U \rightarrow W \rightarrow 0$$
of $k$-vector spaces and any map of $k$-vector spaces $\psi: V \rightarrow W$ there is a (non-unique) map of $k$-vector spaces $\phi: V \rightarrow U$ with
$$\rho \circ \phi = \psi.$$
If you have an exact sequence of $k$-vector spaces
$$0 \rightarrow W \rightarrow U \rightarrow^p U/W \rightarrow 0$$
it follows there is a section $s$ of $p$: There is a map of $k$-vector spaces $s: U/W \rightarrow U$ with $p \circ s = Id$ is the identity map. When you choose a section $s$ you choose a complement to $W$ in $U$ in the following sense: Let $u:=s \circ p$. It follows
$$u^2:= u \circ u = s \circ p \circ s \circ p = s \circ p =u$$
hence $u: U \rightarrow U$ is an idempotent endomorphism. From this it follows you may write
$$U \cong W \oplus Im(u).$$
Hence choosing a section $s$ you get a complement $Im(u)$ to the subspace $W \subseteq U$.
Given any section $s$ of $p$ and any $k$-linear map $v:U/W \rightarrow W$, it follows the new map
$$s+v: U/W \rightarrow U$$
is another section of $p$. The set of sections of $p$ is "parametrized" by the vector space
$$Hom_k(U/W,W).$$
Hence the "set of complements of $W$ in $U$" is a large set in general: There is no unique complement.
Question: "Answers which give an example of such a construction and/or mention applications in which such a construction is useful would be very much appreciated!"
Example: If $G:=SL(V)$ where $V:=\mathbb{C}\{e_1,e_2\}$ and if $U$ is any finite dimensional $G$-module that is not irreducible you may always find a non-zero irreducible sub-$G$-moule $W \subseteq U$. You get an exact sequence of modules
$$S1.\text{ }0 \rightarrow W \rightarrow U \rightarrow U/W \rightarrow 0.$$
The sequence $S1$ always (by the above argument) splits as a sequence of vector spaces. Hence you may always write $U \cong W \oplus Im(u)$ for some idempotent endomorphism $u$. In this case - since $G$ is a semi simple algebraic group you may choose the complement to be a $G$-submodule: You may choose a complement
$$D1.\text{ }U \cong W \oplus W'$$
where $W' \subseteq U$ is a sub-$G$-module. Groups such as $SL(V)$ arise in mathematics and mathematical physics. In many applications you need to calculate such a decomposition as in $D1$. By induction you get a decomposition
$$ U \cong W_1\oplus \cdots \oplus W_k$$
where $W_i$ is an irreducible $G$-module.
In general given any partition
$$\lambda_n \leq \lambda_{n-1} \leq \cdots \leq \lambda_1$$
you get an irreducible $SL(\mathbb{C}^n)$-module denoted $\mathbb{S}_{\lambda}(\mathbb{C}^n)$. Given another partition $\mu$ you get the module
$$W:=\mathbb{S}_{\mu}(\mathbb{S}_{\lambda}(\mathbb{C}^n)),$$
and the module $W$ is not irreducible in general. It is an open problem to decompose $W$ into its irreducible components.
Example: Let $V:=\mathbb{C}^n$ and consider the canonical map
$$ m:V \otimes_k \wedge^k V \rightarrow \wedge^{k+1} V$$
It follows $ker(m) \subsetneq V\otimes_k \wedge^k V$ is an irreducible strict sub-module, hence $V\otimes_k \wedge^k V$ is not irreducible.