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.
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.