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Let's define :

  1. $PY:= \left\lbrace \gamma : I \longmapsto Y : \gamma(0) = y_0 \right\rbrace$

  2. given $f : X \longmapsto Y$ the homotopic fiber $F(f) := \left\lbrace (x,\gamma) \in X \times PY : f(x) = \gamma(1) \right\rbrace$.

  3. The space loop on $X$ $\Omega X := \text{Hom}(\mathbb{S}^1,X)$

The following Lemma is needed in order to arrive to the long h-exact sequence in homotopy

$$\cdots \longmapsto \Omega^2 F(f) \longmapsto \Omega^2 X \longmapsto \Omega^2 Y \longmapsto \Omega F(f) \longmapsto\Omega X \longmapsto \Omega Y \longmapsto F(f) \longmapsto X \longmapsto Y$$

Where $f^1 : F(f) \longmapsto X$ takes $(x,\gamma) \longmapsto x$

Lemma : There's an homeomorphism $\tau^1 : f(\Omega f) \longmapsto \Omega F(f)$ such that $\Omega (f^1) \circ \tau^1 = (\Omega f)^1$.

I didn't find any reference or proof to this fact, except the hint that I should use the following :

Proposition : If $X,Y,Z$ are Hausdorff pointed spaces with $Y$ locally compact, exists a bijection which is also an homeomorphism :

$$\text{Hom}(X \wedge Y, Z)^0 \longmapsto > \text{Hom}(X,\text{Hom}(Y,Z))$$

Any help or reference would be appreciated, I think the problem is proving continuity of the function one will define.

juan diego rojas
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jacopoburelli
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1 Answers1

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Tyrone's comment is the main point. The proposition implies that $P(\Omega X) \cong \Omega(PX)$ by letting the first two variables be $S^1$. In essence, this is just "interchanging variables".

$F(\Omega f)$ is the fiber product $P(\Omega Y)\times_{\Omega Y}\Omega X$ of $P(\Omega Y)$ with $\Omega X$ along the standard maps. The homeomorphism $$ F(\Omega f) = P(\Omega Y)\times_{\Omega Y}\Omega X \cong \Omega(PY) \times_{\Omega Y} \Omega X$$ follows by checking that under the identification $P(\Omega Y) \cong \Omega(PY)$ the maps $P(\Omega Y) \to \Omega Y$ and $(\Omega(PY) \to \Omega Y) = \Omega(PY \to Y)$ coincide. By functoriality of $\Omega$, we have that $\Omega(Ff) = \Omega(PY\times_Y X) \cong \Omega(PY)\times_{\Omega Y}\Omega X$ and the Lemma follows.

The map $\tau^1$ can be made explicit by following the homeomorphisms above. I can include it here, but I recommend that you do it yourself.

juan diego rojas
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