8
$\begingroup$

The question is about a fact that is mentioned as "evident" everywhere in the literature, so my guess is that some small detail is passing over my head. Here it is:

Let $X_0\hookrightarrow X_1 \hookrightarrow X_2 \hookrightarrow \cdots$ be a sequence of inclusions of Hausdorff spaces. Then $$\Omega\left( \varinjlim X_k\right)=\varinjlim\Omega X_k.$$

This is my work so far: For each $k$ we have an obvious inclusion $i_k: \Omega X_k \to \Omega\left( \varinjlim X_k\right)$ given by expanding the codomain of an element $f:\mathbb S^1\to X_k$ to all of $\varinjlim X_k$, and those inclusions induce a continuous injective map $\varphi:\varinjlim\Omega X_k\to \Omega\left( \varinjlim X_k\right)$. A compactness argument shows that the image of $\mathbb S ^1$ under an element of $\Omega\left( \varinjlim X_k\right)$ should lie in one of the $X_k$, which implies that $\varphi$ is actually surjective.

The part I'm stuggling with is showing that $\varphi$ is actually a homeomorphism, i.e. showing that it is an open mapping. I have tried looking explicitly at the compact-open topology in the loop spaces but it seems too hairy.

Any ideas about how to complete this argument or some other different approach would be appreciated.

$\endgroup$
2
  • $\begingroup$ You mean pointed inclusions of pointed Hausdorff spaces,right? $\endgroup$ Commented May 2, 2017 at 7:31
  • 3
    $\begingroup$ I guess, you might need closed inclusions even for the bijection. For the homoemorphism: If we work in compactly generated spaces, it is enough to show that for every compact pointed $L$, the map $\phi$ induces a bijection from maps from $L$. This should follow by the same argument you gave. $\endgroup$ Commented May 2, 2017 at 8:36

0

You must log in to answer this question.