Structure and topology of the abelianization of $GL(n)$ I'm interested in the structure and topology of the abelianization
$$GL(n)^{ab} := GL(n) /[GL(n), GL(n)]$$
of the general linear group $GL(n,\mathbb{R})$, that is, its quotient  by its commutator subgroup.
By using $\det$, it can be shown that this commutator is $SL(n)$. Since $\ker \det = SL(n)$, the first isomorphism theorem thus gives $GL(n)^{ab} \cong \mathbb{R}^{\times}$.
I'm trying to see if I can derive $\det$ up to a constant by showing that the canonical projection $\pi: GL(n) \to GL(n)^{ab}$ extends to a multi-linear and anti-symmetric map $\overline{\pi} : M_{n\times n} \to \overline{GL(n)^{ab}}$ in the columns of matrices, so I'm avoiding the usage of $\det$ in the process.
Is there an alternative way to show that $GL(n)^{ab}$ is isomorphic / homeomorphic to $\mathbb{R}^{\times}$, which does not involve using $\det$ ?
 A: Yes.  Show that $[GL(n,\mathbb{R}),GL(n,\mathbb{R})]$ contains all elementary matrices $E_{ij}(\alpha)$ (same as identity matrix except an $\alpha\in\mathbb{R}$ at $i,j$ entry) and show that every matrix in $SL(n,\mathbb{R})$ is a product of elementary matrices.  Finally, show that every elementary matrix is in the commutator subgroup.
Sketch: Suppose $A\in SL(n,\mathbb{R})$.  The first column is not all $0$, so left multiplying by some $E_{1i}(1)$ we may assume $A_{11}\neq 0$.  Then clear the first column and the first row by multiplying by $E_{i1}(\star)$ on the left and $E_{1i}(\star)$ on the right.  This kills all the off-diagonal terms, and for the diagonal you are reduced to showing $\begin{pmatrix}u\\&u^{-1}\end{pmatrix}$ is a product of elementary matrices.  Finally, note that
$$
E_{ij}(\alpha)=[E_{ir}(\alpha),E_{rj}(1)]
$$
(using the convention $[a,b]=aba^{-1}b^{-1}$ here).
A: This is more an extended comment than a real answer.
The map $\pi : GL(n) \to GL(n)^{ab}$ is a homomorphism of groups. It is not at all clear how it is related to the linear structure of the "column factors" $C_i \approx \mathbb R^n$ of $M(n) = C_1 \times \cdots  \times C_n$. Moreover, it is a priori not clear what $GL(n)^{ab}$ looks like.
In order to be successful you have to do the following:
(1) Construct a group isomorphism $\phi: GL(n)^{ab} \to \mathbb R^\times$
(2) Construct a map $\Pi : M(n) \to \mathbb R$ which extends $\phi \circ \pi$
(3) Show that $\Pi$ is multilinear and antisymmetric.
Step (1) is needed to give a meaning to $\overline{GL(n)^{ab}}$. Note that you must construct $\overline{GL(n)^{ab}}$ to be a one-dimensional linear space in order to give $\overline{\pi}$ a chance to be a multilinear form.
I am sceptical whether this strategy (if it works) is worth to avoid the use of $\det$. It is clear that $\Pi$ must have the form $\Pi = c\det$ with some $c \in \mathbb R$, and constructing $\Pi$ would be an alternative way to introduce $\det$.
