Exercise 2.25 of symplectic topology by McDuff and Salamon asks me to prove that $O(2n)/U(n)$ is homotopy equivalent to $GL(2n,\mathbb{R})/GL(n,\mathbb{C})$. They suggest to use the polar decomposition. I guess they mean something like this: if $A=PQ$, with $P$ positive definite, $Q$ orthogonal, then $P=\sqrt{AA^T}$, and $Q=(\sqrt{AA^T})^{-1}A$, so this decomposition is continuous in $A$. The obvious thing to try is then
$$\Psi_t(A)=(AA^T)^{-t/2}A$$
This retracts $GL(2n,\mathbb{R})$ to $O(2n)$. However, if I multiply $A$ by a matrix $B$, I have
$$ \Psi_t(AB)=(ABB^TA^T)^{t/2}AB $$
Now if $B$ is in $U(n)$ then I see that $\Psi_t(AB)=\Psi_t(A)B$ as $BB^T=1$, so it passes to a homotopy equivalence of the quotients $O(2n)/U(n)$ and $GL(2n,\mathbb{R})/U(n)$. This argument does not work for $B\in GL(n,\mathbb{C})$. What did McDuff and Salamon have in mind?
ps: Note that I am not looking for another proof of the statement (see also Inclusion $O(2n)/U(n)\to GL(2n,\mathbb{R})/GL(n,\mathbb{C}) $). I think the following argument works: We have a diagram of fibrations
$$\begin{array}{ccc} U(n) & \rightarrow & O(2n) &\rightarrow&O(2n)/U(n)\\ \downarrow & & \downarrow & & \downarrow & \\ GL(n,\mathbb{C}) & \rightarrow & GL(2n,\mathbb{R}) & \rightarrow & GL(2n,\mathbb{R})/GL(n,\mathbb{C}) \end{array}$$
The first two vertical maps are homotopy equivalences. Now the long exact sequence of homotopy groups, the five lemma, and Whitehead will finish the job. But I am looking for an explicit homotopy equivalence.
Basically, the homotopy equivalence you're after is just the inclusion.
Let $ O(2n) \to GL_{2n}(\mathbb R)$ be the inclusion. If you compose it with the canonical surjection, you get a well-defined continuous map $O(2n) \to GL_{2n}(\mathbb R)/GL_n(\mathbb C).$ Because $U(n)$ is a subgroup of $GL_n(\mathbb C)$, multiplying an element of $O(2n)$ by a matrix in $U(n)$ doesn't change its image, so this defines a continuous map $O(2n)/U(n) \to GL_{2n}(\mathbb R)/GL_n(\mathbb C)$. It's really as explicit as it gets. As I said, you can really think of this map as the inclusion.
I think that you were simply confused by McDuff and Salmon's hint. But they were probably just reminding you of the polar decomposition which is the key to prove that the inclusions $U(n) \to O(2n)$ and $GL_n(\mathbb C) \to GL_{2n}(\mathbb R)$ are homotopy equivalences: the argument you hinted at in your question is really what this exercise is about.