Homotopic type of $GL^+(n)$, $SL(n)$ and $SO(n)$ Question:

Consider $GL^+(n) \supset SL(n) \supset SO(n)$ the groups of matrices $n \times n$ with positive determinant, determinant $1$ and orthogonal with positive determinant, respectively. Show that these three spaces have the same homotopy type. 

Attempt: Two spaces are said to have the same homotopy type when for a continuous map $f : X \to Y$ there exists a continuous $g : Y \to X$ such that $f \circ g \simeq id_Y$ and $g \circ f \simeq id_X$. 
(Main idea): Now if we show that $GL^+(n)$ is homeomorphic to $\mathbb R^+ \times SL(n)$ and $P \times SO(n)$, where $P$ is the set of all positive definite matrices in $GL^+(n)$, then we are done. 
For the second homeomorphism we notice that for any matrix $A \in GL^+(n)$ if we take the positive definite matrix $A^TA$ then we may find a unique positive definite matrix $B$ such that $B^2 = A^T A$. Then $AB^{-1}$ is orthogonal and we have the (unique) decomposition $A = AB^{-1}\cdot B$. We consider the map $$\begin{align} \psi: P \times SO(n) &\to GL^+(n)\\(X,A) &\mapsto A\exp X\end{align}$$
is a homeomorphism (still have to show that is surjective). 
As for the first homeomorphism I still haven't figured out. I thought maybe considering 
$$\begin{align} \varphi: \mathbb R^+ \times SL(n) &\to GL^+(n)\\(t,A) &\mapsto \exp tA\end{align}$$
or maybe something like 
$$\begin{align} \varphi: Gl^+(n) &\to\mathbb R^+ \times SL(n) \\A &\mapsto (\det A, \,\,\,?\,\,\,)\end{align}$$
Any ideas are welcome. 
Note: I am not interested in any different approaches other than the one presented. 
 A: Your approach for the first question will work, but you need to note that the matrix $B$ you construct necessarily has distinct eigenvalues: since $A^TA$ is symmetric and positive definite, it has distinct eigenvalues, and thus so does its square root.  (This is how you contract your space $P$ by the way - send the eigenvalues - which are positive reals - continuously to 1.  The same argument does not work over the complex numbers, by the way!)
You should also be a little careful with matrix square roots and continuity.  But I'll leave you to solve that problem.

For the second decomposition things are much simpler.  Remember that $$\det(tA) = t^n\det(A)$$ for any $t \in \mathbb R$ and $A \in GL(n)$.  So let's construct the maps $$f: GL^+(n) \to \mathbb R^+ \times SL(n), \qquad f(A) = \left(\sqrt[n]{\det A},\frac{1}{\sqrt[n]{\det A}}A\right)$$ and $$g: \mathbb R^+ \times SL(n) \to GL(n), \qquad g(t, B) = tB.$$  Then clearly $f$ and $g$ are inverses; they're both continuous and hence give the desired homeomorphisms.
