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Section 10.5 HELP for a Disc via Fibrations

We now prove the disc case of HELP promised in Lemma 8.8. The idea is the same as in the homotopy-fibre argument: an \(n\)-equivalence has \((n-1)\)-connected homotopy fibres, so a map from \(S^{n-1}\) into such a fibre extends over \(D^n\text{.}\)

Definition 10.20.

Given \(e\colon Y\to Z\text{,}\) the mapping path space \(P_e\) (or mapping cocylinder) is the pullback of \(e\) along \(\mathrm{ev}_0\colon Z^I\to Z\text{,}\) i.e.,
Explicitly, \(P_e=\{(y,\gamma)\in Y\times Z^I:\gamma(0)=e(y)\}\text{,}\) where path spaces carry the compact-open topology. Note that \(P_e\simeq Y\) (Lemma 10.22).

Proof.
Our test diagram is the following:
Note that \(g(a)\in P_e\subseteq Y\times Z^I\text{,}\) so we can write \(g(a)=(g_1(a),g_2(a))\text{.}\) Here \(g_1(a)\) maps via \(e\) to the starting point of the path \(g_2(a)\text{,}\) and the commutativity of the diagram implies that the endpoint of the path \(g_2(a)\) is the starting point of \(H(a,-)\text{.}\) The lift \(\widetilde H\) will have two components. The \(Y\)-component will be constant in \(s\text{,}\) i.e., \(\widetilde H_1(a,s)=g_1(a)\text{.}\) Overall, we define
\begin{equation*} \widetilde H(a,s)=\bigl(g_1(a),\widetilde H_2(a,s)(-)\bigr)\in P_e, \end{equation*}
where
\begin{equation*} \widetilde H_2(a,s)(t)= \begin{cases} g_2(a)\bigl((1+s)t\bigr) & 0\leq t\leq 1/(1+s),\\ H\bigl(a,(1+s)t-1\bigr) & 1/(1+s)\leq t\leq 1. \end{cases} \end{equation*}
One checks directly that \(\widetilde H(a,s)\) has the required properties.

Proof.
We have \(qj=\mathrm{id}_Y\text{,}\) and shrinking paths by \(\gamma(t)\mapsto\gamma((1-s)t)\) gives \(\mathrm{id}_{P_e}\simeq jq\text{.}\) Finally \(pj(y)=c_{e(y)}(1)=e(y)\text{.}\)

Definition 10.23.

Let \(e\colon Y\to Z\) and \(z\in Z\text{.}\) The homotopy fibre of \(e\) over \(z\) is the fibre of the fibration \(p\colon P_e\to Z\) of Proposition 10.21,
\begin{equation*} F_e(z)=p^{-1}(z)=\{(y,\gamma)\in Y\times Z^I:\gamma(0)=e(y),\ \gamma(1)=z\}. \end{equation*}
Thus a point of \(F_e(z)\) consists of \(y\in Y\) together with a path from \(e(y)\) to \(z\text{.}\) When \(z\) is understood we simply write \(F_e\text{.}\)
Equivalently, let \(P_zZ=\{\gamma\in Z^I:\gamma(1)=z\}\) be the space of paths ending at \(z\text{.}\) The condition \(\gamma(1)=z\) says exactly that \(\gamma\in P_zZ\text{,}\) so \(F_e(z)=\{(y,\gamma)\in Y\times P_zZ:\gamma(0)=e(y)\}\text{.}\) That is, \(F_e(z)\) is the pullback

Proof.
By Lemma 10.22, \(pj=e\) with \(j\) a homotopy equivalence, so \(p\) is an \(n\)-equivalence whenever \(e\) is.
Surjectivity on path components makes every fibre nonempty. For \(1\leq k\leq n-1\text{,}\) the segment
\begin{equation*} \pi_{k+1}(P_e)\longrightarrow\pi_{k+1}(Z) \longrightarrow\pi_k(F_e(z))\longrightarrow\pi_k(P_e) \longrightarrow\pi_k(Z) \end{equation*}
of the fibration sequence has its last map an isomorphism, since \(k<n\text{,}\) and its first map surjective, since \(k+1\leq n\) (it is even an isomorphism unless \(k=n-1\)). By exactness, the boundary map \(\pi_{k+1}(Z)\to\pi_k(F_e(z))\) is zero and so is \(\pi_k(F_e(z))\to\pi_k(P_e)\text{,}\) hence \(\pi_k(F_e(z))=0\text{.}\)
We leave it as an exercise to show that the fibre is also path-connected, which includes the case \(n=1\text{.}\) Hint: use that \(p\) is injective on \(\pi_0\) and surjective on \(\pi_1\text{.}\)

Remark 10.25.

We now make a series of remarks about a map \(f\colon X\to Y\) with homotopy fibre \(F_f\) over \(y_0 \in Y\) (Definition 10.23).
  1. A map \(\phi\colon S^{n-1}\to F_f\) corresponds to a commutative diagram
    \begin{equation} \begin{CD} S^{n-1} @>{g}>{}> X \\ @V{}V{}V @V{}V{f}V \\ C(S^{n-1})\cong D^n @>{}>{h}> Y \end{CD}\tag{10.1} \end{equation}
    where \(g\) is the composite \(S^{n-1}\xrightarrow{\phi}F_f\to X\text{,}\) \(h\) is adjoint to the path coordinate of \(\phi\text{,}\) and the cone point goes to \(y_0\text{.}\)
    To see this, recall from Definition 10.23 that \(F_f\) is the pullback of \(f\) along \(\mathrm{ev}_0\colon P_{y_0}Y\to Y\text{.}\) Maps into a pullback form a pullback of sets. Moreover, the adjunction of Remark 7.12, restricted to paths ending at \(y_0\text{,}\) identifies maps \(S^{n-1}\to P_{y_0}Y\) with maps \(C(S^{n-1})=S^{n-1}\times I/S^{n-1}\times\{1\}\to Y\) sending the cone point to \(y_0\text{.}\) Hence
    \begin{align*} \Hom(S^{n-1},F_f) &\cong \Hom(S^{n-1},X)\times_{\Hom(S^{n-1},Y)}\Hom(S^{n-1},P_{y_0}Y)\\ &\cong \Hom(S^{n-1},X)\times_{\Hom(S^{n-1},Y)}\Hom_*(D^n,Y), \end{align*}
    where \(\Hom_*(D^n,Y)\) consists of the maps sending the centre of \(D^n\cong C(S^{n-1})\) to \(y_0\text{,}\) and \(\Hom_*(D^n,Y)\to\Hom(S^{n-1},Y)\) is restriction to the boundary. An element of the right-hand side is a pair \((g,h)\) with \(h|_{S^{n-1}}=f\circ g\text{,}\) which is exactly the commutative square (10.1).
    In other words, a map into \(F_f\) is a map \(g\) into \(X\) together with a choice of null-homotopy \(h\) of \(fg\) ending at \(y_0\text{.}\) The null-homotopy is data carried by \(\phi\text{,}\) not a property of \(g\text{:}\) a given \(g\) may admit no such \(h\text{,}\) or many, and different choices of \(h\) give different maps \(\phi\text{.}\)
  2. The boundary map \(\pi_n(Y)\to\pi_{n-1}(F_f)\) in the long exact sequence corresponds to the map sending the class of \(\bar h\colon S^n\to Y\) to the class of \(\pi_{n-1}(F_f)\) represented by the diagram
    where \(c=c_{x_0}\) is the constant map, and \(h\) is the composite \(D^n\to D^n/S^{n-1}\cong S^n\xrightarrow{\bar h}Y\text{.}\)
  3. Similarly, the map \(\pi_{n-1}(F_f)\to\pi_{n-1}(X)\) corresponds to sending the diagram (10.1) to the class \([g]\text{.}\)
  4. In particular, \(\pi_{n-1}(F_f)=0\) is equivalent to completing the diagram (10.1) in the following way:
We can restate the last remark in the following lemma.

Proof.
For any disk we have a homeomorphism \(CD^n\cong D^n\times I\) which sends the cone point to the center of \(D^n\times\{1\}\text{,}\) \(D^n\) to \(S^{n-1}\times I\cup D^n\times\{0\}\text{,}\) \(S^{n-1}\) to \(S^{n-1}\times\{1\}\) and \(CS^{n-1}\) to \(D^n\times\{1\}\text{.}\) Thus the statement follows from the last part of the remark.

Proof.
Proof of Lemma 8.8. Let \(n\geq1\) and let \(e\colon Y\to Z\) be an \(n\)-equivalence. The HELP data for \((D^n,S^{n-1})\) is exactly the first diagram of Lemma 10.26, with \(e\colon Y\to Z\) in place of \(f\colon X\to Y\text{:}\) the map \(g\) on \(S^{n-1}\times\{1\}\text{,}\) and the given maps \(D^n\to Z\) and \(h\) together on \(S^{n-1}\times I\cup D^n\times\{0\}\text{.}\) The required extensions \(\widetilde g\) and \(\widetilde h\) are the maps \(G\) and \(H\text{.}\) By Lemma 10.24 every homotopy fibre of \(e\) is \((n-1)\)-connected, so Lemma 10.26 applies. For \(n=0\text{,}\) surjectivity on \(\pi_0\) supplies a point of \(Y\) and a path from the given point of \(Z\) to its image.

Remark 10.27. The path-space fibration.

For \(e\colon\{z_0\}\hookrightarrow Z\text{,}\) the mapping path construction gives
\begin{equation*} \Omega Z\longrightarrow PZ\xrightarrow{\mathrm{ev}_1}Z, \end{equation*}
where \(PZ\) consists of paths starting at \(z_0\) and \(\Omega Z\) consists of loops at \(z_0\text{.}\) Shrinking paths contracts \(PZ\text{,}\) so the long exact sequence gives \(\pi_k(\Omega Z)\cong\pi_{k+1}(Z)\) for \(k\geq1\text{;}\) for \(k=0\text{,}\) path components of \(\Omega Z\) are exactly based homotopy classes of loops. We will use this fibration repeatedly in the spectral sequence computations.
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