Inverse limit of sequence of fibrations

Question

I'm reading through the proof of Proposition 4.67 in Hatcher's Algebraic Topology, and I've come to something that I'm having trouble understanding.

For an arbitrary sequence of fibrations $... \rightarrow\ X_2 \rightarrow\ X_1$, I need to show that the natural map $\lambda: \pi_i(\varprojlim X_n) \rightarrow\ \varprojlim \pi_i(X_n)$ is injective if the maps $\pi_{i+1}(X_n) \rightarrow\ \pi_{i+1}(X_{n-1})$ are surjective for all $n$.

If $f: S^i \rightarrow\ \varprojlim X_n$ is in $Ker \lambda$, then each component map $f_n: S^i \rightarrow\ X_n$ is nullhomotopic. So for each $n$, there exists $F_n: D^{i+1} \rightarrow\ X_n$ such that $F_n$ restricted to $S^i$ is $f_n$. If $F_n$ can be homotoped so that $F_{n-1} = p_nF_{n}$, where $p_n: X_n \rightarrow\ X_{n-1}$ is from the sequence of fibrations, then induction on $n$ yields the result.

From the definition of the inverse limit, $F_{n-1} = p_nF_{n}$ on $S^i$, so by gluing the two disks $p_nF_n: D^{i+1} \rightarrow\ X_{n-1}$ and $F_{n-1}: D^{i+1} \rightarrow\ X_{n-1}$ together along $S^i$, it is possible to construct a map $g_{n-1}: S^{n+1} \rightarrow\ X_{n-1}$.

According to Hatcher, the surjectivity of $\pi_{i+1}(X_n) \rightarrow\ \pi_{i+1}(X_{n-1})$ implies that there exists $G_n: D^{i+1} \rightarrow\ X^n$ such that $G^n$ restricted to $S^i$ is $f_n$ and $p_nG_n$ is homotopic rel $S^i$ to $F_{n-1}$.

Why is this true? If $\pi_{i+1}(X_n) \rightarrow\ \pi_{i+1}(X_{n-1})$, then there exists $g_{n}: S^{i+1} \rightarrow\ X_{n}$ such that $p_ng_n$ is homotopic to $g_{n-1}$, but I see no way to get the desired $G_n$ from $g_n$. Any help would be most appreciated.

Answer 1

To be honest, I have trouble finding out what is the needed argument, but I'll try to give another one. I will consistently use radial coordinates on the disk, ie. $(r, \phi) \in D^{k}$, where $r$ is the distance from the centre and $\phi \in S^{k-1}$.

Let $\tilde{F}_{n-1}: S^{i} \times I \rightarrow X_{n-1}$ given by $\tilde{F}_{n-1}(\phi, t) = F_{n-1}(t-1, \phi)$ be the contracting homotopy. By the fibration property there exists a lift of this homotopy to $X_{n}$, say $G_{n}$, such that $G_{n} |_{S^{i} \times \{ 0 \}} = f_{n}$.

Let $g_{n} = G_{n} |_{S^{i} \times \{ 1 \}}$. Since $\tilde{F}_{n-1} |_{S^{i} \times \{ 1 \}}$ maps to the basepoint, we see that the image of $g_{n}$ is contained in the fibre $F \subseteq X_{n}$. The map $\pi_{i+1}(X_{n}) \rightarrow \pi_{i+1}(X_{n-1})$ is surjective and hence by the long exact sequence of homotopy the map $\pi_{i}(F) \rightarrow \pi_{i}(X_{n})$ is injective. Since $g_{n}$ is null in $\pi_{i}(X_{n})$ (being homotopic to $f_{n}$), it must be also null in $F$ and thus there is a contracting homotopy $H_{n}: S^{i} \times I \rightarrow F$.

Now define a map $F ^\prime _{n}: D^{i+1} \rightarrow X_{n}$ by

$F ^\prime _{n}(r, \phi) = G_{n}(\phi, 2(1-r))$ for $r \geq \frac{1}{2}$

$F ^\prime _{n}(r, \phi) = H_{n}(\phi, 2(\frac{1}{2}-r))$ for $r \leq \frac{1}{2}$,

After a moment of thought, we see that $p_{n} F^\prime _{n}$ is homotopic to $F_{n-1}$. Using the homotopy lifting for $(D^{i+1}, S^{i})$ we may deform it $rel \ S^{i}$ to a map $F_{n}$ such that $p_{n} F_{n} = F_{n-1}$, which is what we wanted to do.

Answer 2

I believe I have figured something along the lines of Hatcher's proof! I have been able to construct this $G_n$ such that $p_nG_n \simeq F_{n-1}$ but this homotopy does not fix $S^i$. However, we can still get by without it. We will use the following claim.

claim: For any $[\phi] \in \pi_q(Y,y_0)$ and for any map $f \colon (D^q, *) \to (Y,y_0)$, there is a map $f' \colon (D^q, *) \rightarrow (Y,y_0)$ such that $f|_{S^{q-1}} = f'|_{S^{q-1}}$ and $ [f\cup f']=[\phi] $, where $f \cup f' \colon (D^q \cup_{S^{q-1}} D^q , *) \to (Y,y_0).$

proof of claim: Consider $D^q$ as the upper hemisphere $S^q_+$, so that $f \colon S^q_+ \to Y$. Define a map on the lower hemisphere $\bar f \colon S^q_- \to Y$ by projecting vertically to the upper hemisphere and then going through $f$, i.e. by the formula $$\bar f(x_1 , \dots , x_q) := f(x_1, \dots , - x_q).$$ Then the map $f\cup \bar f \colon S^q \to Y$ extends to a map of the disk, so it is null homotopic. Then we have $[\phi]+[f \cup \bar f] = [\phi] $. Recall that the sum of these classes is represented by, the composition $$S^q \stackrel{p}{\longrightarrow} S^q \vee (S^q_+ \cup S^q_-) \stackrel{\phi \vee (f\cup \bar f)}{\longrightarrow} Y $$ We will then identify $D^{q} $ with $p^{-1}(S^q\vee S^q_-)$ and define $f'$ to be the restriction of the above composition to $D^q$. end of proof ///

Once we constructed this map $g_{n-1} \colon S^{i+1} \to X_{n-1}$, we use the surjectivity of $(p_n)_* \colon \pi_{i+1}(X_n) \to \pi_{i+1}(X_{n-1})$ to lift $[g_n]$ to a class $[g_n] \in \pi_{i+1}(X_n)$. We then apply the claim to this class and the original null homotopy $F_n : D^{i+1} \to X_n$, to obtain a map $G_n \colon D^{i+1} \to X_n$ such that $F_n |_{S^i} = G_n |_{S^i}$ and $[F_n \cup G_n] = [g_n]$. This map $G_n$ will be our new null homotopy. Observe that $(p_n)_* ([F_n \cup G_n]) = [p_nF_n \cup p_nG_n] = [g_n].$ But recall that $g_n= p_nF_n \cup F_{n-1}$, so there is a homotopy $$H\colon (D^{i+1} \cup_{S^{i}}D^{i+1})\times I \to X_{n-1}$$ which restricts on the second hemisphere to a homotopy from $p_nG_n$ to $F_{n-1}$.

Now we will see that we don't actually need this homotopy to fix the circle (although if anyone sees a way to force it to, that would be cleaner). Using the lifting property of $p_n$, we obtain a map $K_n \colon D^{i+1} \to X_n$ such that $p_nK_n = F_{n-1}$ and $f_n \simeq K_n|_{S^{i}}$ . Lets use $k_n$ to denote this restriction. The issue now is that the components $\{ \dots ,f_{n+1}, k_n , f_{n-1}, \dots, f_1 \}$ no longer define a map $S^i\rightarrow \varprojlim X_i$ since $p_{n+1} f_{n+1}\neq k_n$. But these maps are homotopic, and that is good enough since we can replace $f_{n+1}$ with a homotopic map $k_{n+1}$ using the lifting property of $p_{n+1}$ so that $p_{n+1} k_{n+1} = k_n$. Letting induction roll this process out to infinity, we have created a map $k \colon S^{i} \rightarrow \varprojlim X_n$ which is homotopic to our original map $f$, which extends to a $K : D^{i+1} \rightarrow \varprojlim X_n$.