I'm reading Bott and Tu's book "Differential forms in Algebraic Topology" and I need some help to understand a detail on the long exact sequence (LES) of homotopy groups for a (Hurewicz or Serre) fibration.
Given a base point preserving fibration $p : E \to B$, with fiber $F \hookrightarrow E$, there is a LES of homotopy groups:
$$\pi_q(F) \xrightarrow{i_*} \pi_{q}(E) \xrightarrow{p_*} \pi_q(B) \xrightarrow{\partial} \pi_{q-1}(F) \to \cdots$$
I would like to understand the construction of the boundary map from Bott and Tu's perspective. Their argument goes as follows (if i'm correct):
- An element $\alpha \in \pi_q(B)$ is identified with a map from the $q$-cube $\alpha: I^q \to B$ such that $\alpha(\partial I^q)= \ast_B$, where $\ast_B$ is the base point and $\partial I^q$ is the boundary.
- Using the embeddings $I^{q-1} \cong I^{q-1} \times \{0\} \hookrightarrow I^q$ and $I^{q-1} \cong I^{q-1} \times \{1\} \hookrightarrow I^q$ , we can view $\alpha : I^{q-1} \times I \to B$ as a homotopy between two (constant ?) maps $\alpha|I^{q-1} \to B$. I'm guessing these two maps are indeed constant since $I^{q-1} \times \{0\} \subset \partial I^q$ and $I^{q-1} \times \{1\} \subset \partial I^q$.
- The constant map $I^{q-1} \times \{0\} \to E$ of value $\ast_E$ covers the constant map $\alpha_{|I^{q-1} \times \{0\}} : I^{q-1} \to B$ of value $\ast_B$, and we can use the covering homotopy property to find a homotopy $\tilde{\alpha} : I^{q-1} \times I \to E$ satisfying:
$$ p \circ \tilde{\alpha} = \alpha \quad \quad \text{and} \quad \tilde{\alpha} (I^{q-1} \times \{0\}) = \ast_E$$ - The equality $p \circ \tilde{\alpha} (I^{q-1} \times \{1\}) = \alpha(I^{q-1} \times \{1\}) = \ast_B $ implies that $\tilde{\alpha}(I^{q-1} \times \{1\}) \subset p^{-1}(\ast_B) = F$
- Then they define $\partial [\alpha]$ as the homotopy class of (this map ?) $\tilde{\alpha}:(t_1, \cdots, t_{q-1}, 1) \to F$. This is where I don't get it !
How does $\tilde{\alpha}:(t_1, \cdots, t_{q-1}, 1) \to F$ define an element
in $\pi_{q-1}(F)$ since we don't know yet if this map is constant (of value $\ast_E$) on the boundary of $I^{q-1}$ ?
Maybe I'm missing something in this proof. Can someone help ?
I'm adding a screenshot of the paragraph (if that's okay).
Boundary map explained
Best Answer
First, here is a relatively explicit homeomorphism from $(I^n, I^{n-1}\times \{0\})$ and $I^n, I^{n-1}\times \{0\} \cup \partial I^{n-1} \times I).$ Actually, I'm going to use $I = [-1,1]$ to make formulas a little bit nicer.
We will view $I^n$ as a union of concentric copies of $\partial I^n$ with a single point at the center. Concretely, for each $t\in [0,1]$, let $I_t:=\{(x_1,...,x_n)\in I^n: |x_i|\leq t$ for every $i$ and $|x_i| = t$ for at least one $i\}$. So, $I_1 = \partial I^n$ and $I_0$ is a single point.
We will define a homeomorphism $f$ of $I_1$ which maps $I^{n-1}\times \{-1\}$ to $I^{n-1}\times \{0\}\cup \partial I^{n-1}\times I$. Then we'll just copy this homeomorphism on each $I_t$.
To begin with, set $p:= (0,...,0,-1)\in I^{n-1}\times \{-1\}$. We set $f(p) = p$.
For every other point $x\in I^{n-1}\times \{-1\}$, there is a unique ray emanating from $p$ to $x$. We define $g(x)$ to be the point where this ray intersects $[-1/2,1/2]^{n-1}\times \{-1\}$ and we let $h(x)$ be where the ray intersects $\partial I^{n-1}\times \{-1\}$.
For $x\in [-1/2,1/2]^{n-1}\times \{-1\}$, we set $f(x) = \left(\frac{d(x,p)}{d(g(x), p)} h(x), -1\right)$, where $d$ is the usual Euclidean distance function. Intuitively, we are radially scaling the smaller cube $[-1/2,1/2]^{n-1}$ to fill the larger cute $I^{n-1}$.
For $x\in I^{n-1}\times \{-1\}$ but outside of $[-1/2,1/2]^{n-1}\times \{-1\}$, we define $f(x) = \left(h(x), \frac{d(x,g(x))}{d(x,h(x))}\right)$. This part surjects onto $\partial I^{n-1} \times I$.
All of this is just the definition of $f$ on the bottom face. So, far, we have a homeomorphism $f:I^{n-1}\times \{-1\}\rightarrow I^{n-1}\times {-1}\cup \partial I^{n-1}\times I$. We want to extend $f$ to $\partial I^{n-1}\times I \cup I^{n-1}\times \{1\}$. However, this domain is obviously homeomorphic to the range of $f|_{I^{n-1}\times \{-1\}}$, we can just use $f^{-1}$ (slightly modified) to extend $f$ to the rest of $I_1$. A little thought will show that this glues together where the two domains overlap.
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How does this help with the connecting homomorphism?
Follow the proof as written until you get to the last step. From the above terrible formulas, $I^{n-1}\times \{1\}$ is homeomorphic to $I^{n-1}\times \{1\}\cup \partial I^{n-1}\times I$. Call such a homeomorphism $f$. Then, instead of declaring $\partial [\alpha] = \tilde{\alpha}(x_1,...,x_{n-1}, 1)$, define $\partial[\alpha] = \tilde{\alpha}(f(x_1,...,x_{n-1},1))$. The point is that $f$ maps the boundary $\partial I^{n-1}\times \{1\}$ onto $\partial I^{n-1}\times \{0\}$, and $\tilde{\alpha} $ has the value $\ast_E$ on that face.