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For a based map $f : X \to Y$, define the "homotopy fiber" $Ff$ to be$$Fd = X \times_f PY = \{(x, \chi) : f(x) = \chi(1)\} \subset X \times PY.$$Equivalently, $Ff$ is the pullback displayed in the diagram

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where $\pi(x, \chi) = x$. As a pullback of a fibration, $\pi$ is a fibration.

If $\rho: Nf \to f$ is defined by $\rho(x, \chi) = \chi(0)$, then $f = \rho \circ \nu$, where $\nu(x) = (x, c_{f(x)})$, and $Ff$ is the fiber $\rho^{-1}(*)$. Thus the homotopy fiber $Ff$ is constructed by first replacing $f$ by the fibration $\rho$ and then taking the actual fiber.

Let $\iota: \Omega Y \to Ff$ be the inclusion specified by $\iota(\chi) = (*, \chi)$. The sequence$$\dots \to \Omega^2 X \overset{\Omega^2 f}{\longrightarrow} \Omega^2Y \overset{-\Omega\iota}{\longrightarrow} \Omega Ff \overset{-\Omega\pi}{\longrightarrow} \Omega X \overset{-\Omega f}{\longrightarrow} \Omega Y \overset{\iota}{\to} Ff \overset{\pi}{\to} X \overset{f}{\to} Y$$is called the fiber sequence generated by the map $f$; here$$(-\Omega f)(\zeta)(t) = (f \circ \zeta)(1 - t) \text{ for }\zeta \in \Omega X.$$These "long exact sequences of based spaces" also give rise to long exact sequences of pointed sets, covariantly.

Theorem. For any based space $Z$, the induced sequence$$\dots \to [Z, \Omega F f] \to [Z,\Omega X] \to [Z, \Omega Y] \to [Z, Ff] \to [Z, X] \to [Z, Y]$$is an exact sequence of pointed sets, or of groups to the left of $[Z, \Omega Y]$, or of Abelian groups to the left of $[Z, \Omega^2Y]$.

Exactness is clear at the first stage. To see this, consider the diagram

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Here $h: c_* \simeq f \circ g$, and we view $h$ as a map $Z \to PY$. Thus we check exactness by using any given homotopy to lift $g$ to the fiber.

I claim that, up to homotopy equivalence, each consecutive pair of maps in my fiber sequence is the composite of a map and the projection from its fiber onto its source. This will imply the source. I observe that, that for any map $f$, interchange of coordinates gives a homeomorphism$$\Omega F f \cong F(\Omega f)$$such that the following diagram commutes:

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Here $\tau$ is obtained by interchanging the loop coordinates and is homotopic to − \text{id}. We have $\iota(f)$, $\pi(f)$, etc., to indicate the maps to which the generic constructions $\iota$ and $\pi$ are applied. Using this inductively, we see that we need only verify our claim for the two pairs of maps $(\iota(f), \pi(f))$ and $(-\Omega f, \iota(f))$.

Anyways, one key step in finishing the proof: I need that the right triangle commutes and the left triangle commutes up to homotopy in the following diagram.

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My question is, what is the easiest way to see that the two triangles commute up to homotopy? Thanks in advance.

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    $\begingroup$ Sorry, what are $X,Y,F,f,\pi,\iota$? $\endgroup$ – Berci Sep 5 '15 at 21:25
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First, we show that the right triangle commutes up to homotopy. Considering the fiber sequence$$F(f) \overset{\pi(f)}{\to} X \overset{f}{\to} Y,$$we want to show that the inclusion of the strict fiber of $\pi(f)$ into its homotopy fiber $F(\pi(f))$ is a homotopy equivalence making the right triangle commute, where$$\iota: \Omega Y \to F(f), \text{ }\iota(\gamma) = (x_0, \gamma) \in F(f) = X \times_Y PY.$$The strict fiber of $\pi(f)$ is the subset of $F(f)$,$$(\pi(f))^{-1}(x_0) = \{(x_0, \gamma) : \gamma(0) = y_0,\,\gamma(1) = f(x_0) = y_0\} \cong \Omega Y,$$and the composite $\Omega Y \to F(\pi(f)) \to F(f)$ is$$\gamma \mapsto (\iota(\gamma),\,\text{constant path at the basepoint of }F(f)) \mapsto \iota(\gamma).$$The inclusion of the strict fiber $\phi: \Omega Y \overset{\simeq}{\to} F(\pi(f))$ is a homotopy equivalence, since $\pi(f)$ is a fibration (and using this).

Now, we show that the left triangle commutes up to homotopy. The homotopy we require is$$H: \Omega X \times I \to F(\pi(f)),\text{ }(\gamma, s) \mapsto ((*, (f \circ \gamma)(1 - t)|_{[0, 1-s]}), \gamma|_{[1 - s, 1]}).$$This is $\phi \circ (-\Omega f)$ at time $0$ and $i \circ \pi(f)$ at time $1$, so we are done.

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