General Linear Group as a Direct Product? Let $K$ be a field and consider the surjective determinant homomorphism $\mathrm{GL}_n(K)\to K^\times$. Since the kernel is the special linear group $\mathrm{SL}_n(K)$ we obtain a short exact sequence
$$\require{AMScd}\begin{CD}
 1 @>>> \mathrm{SL}_n(K) @>i>> \mathrm{GL}_n(K) @>\det>> K^\times @>>> 1\\
\end{CD}$$
Since the determinant map has a section $s:K^\times\to\mathrm{GL}_n(K)$ defined by $$s(\alpha):=\begin{pmatrix} \alpha&&&\\ &1&& \\ &&\ddots& \\ &&&1\end{pmatrix},$$
we conclude from the splitting lemma that $\mathrm{GL}_n(K)\approx\mathrm{SL}_n(K)\rtimes K^\times$. Here's a question:
Under what conditions on $K$ and $n$ does the inclusion homomorphism $i:\mathrm{SL}_n(K)\to\mathrm{GL}_n(K)$ have a retraction?
Example: If $K=\mathbb{R}$ and $n$ is odd, then every $\alpha\in\mathbb{R}^\times$ has a unique real $n$-th root, so we obtain a group homomorphism $\sqrt[n]{\cdot}:\mathbb{R}^\times\to\mathbb{R}^\times$, and we can use this to define a retraction $r:\mathrm{GL}_n(\mathbb{R})\to\mathrm{SL}_n(\mathbb{R})$ by $$r(A):=\frac{1}{\sqrt[n]{\det(A)}}\cdot A.$$
Now it follows from the splitting lemma that $\mathrm{GL}_n(\mathbb{R})\approx \mathrm{SL}_n(\mathbb{R})\times\mathbb{R}^\times$. Here's another question:
Is there a topological/geometric explanation for this example? Is there a topological/geometric obstruction when $n$ is even?
 A: $\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction iff the following two conditions hold


*

*The subgroup of $n$-root of unity in $K^*$ has a direct summand in $K^*$.

*$x\mapsto x^n$ is surjective on $K$


Let us first check that (1.) is equivalent to the existence of a retraction in restriction to $K^*\mathrm{SL}_n(K)$ (the subgroup generated by homotheties and unimodular matrices). Clearly it implies it (consider the set of scalar matrices whose diagonal entry belongs to this direct summand). Conversely, if there is a retraction, its kernel has trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so it should form the set of scalar matrices with diagonal entry in some direct summand of the set $n$-roots of unity in $K^*$.
Now it is clear that $K^*\mathrm{SL}_n(K)$ equals $\mathrm{GL}_n(K)$ if and only if (2.) holds; so if both (1.) and (2.) hold it follows that we have a retraction; conversely if we have a retraction, its kernel is a normal subgroup with trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so $\mathrm{GL}_n(K)$ should be generated by unimodular and scalar matrices, i.e. (2.) holds, and then the first verification shows that (1.) holds.
Edit: as noticed by Julian Rosen, (2.) together with (1.) implies something much stronger than (1.), namely that the subgroup of $n$-roots of unity is actually trivial, which means that $x\mapsto x^n$ is injective on $K$. To conclude:

$\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction (in the category of groups) iff $x\mapsto x^n$ is a permutation of $K$

