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

  • $\begingroup$ I don't have Bott and Tu so I can't see if they mention this, but the usual way of defining the connecting homomorphism is to note that the pair $(I^{n+1}, I^n\times \{0\})$ is homeomorphic to the pair $(I^{n+1}, I^n\times \{0\}\cup \partial I^n \times I)$. So, when you talk about the constant map $I^{q-1}\times \{0\}\rightarrow E$, you could instead interpret is at the constant map from $I^{q-1}\times \{0\}\cup \partial I^{q-1}\times I$. The advantage is that now $\tilde{\alpha}(t_1,..., t_{q-1}) = \ast_E$. $\endgroup$ – Jason DeVito Nov 26 '19 at 3:43
  • $\begingroup$ Thanks for your response. Could you tell me how you finish the proof from here ? Or at least a reference where it's done with details ? Also as you mention it, I couldn't find a reference where the homeomorphism of pairs between $(I^{n+1}, I^n\times \{0\})$ and $(I^{n+1},I^n \times \{0\} \cup \partial I^n \times I)$ is explicit. Everyone keep saying that "it's obvious" but I would like a formula for it (or at least a description). I know how it works for $I^2$ but for higher cubes I don't see it. $\endgroup$ – Mathex Nov 26 '19 at 8:14
  • $\begingroup$ I will try to answer both questions when I have time, gotta go teach for awhile. $\endgroup$ – Jason DeVito Nov 26 '19 at 13:54

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.

$ \ $

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.

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  • $\begingroup$ Here is some intuition for the ugly formulas I wrote, but instead of expanding the bottom face to cover all the sides and the bottom, let's expand the top. Imagine you've painted the top of a hypercube. Then you add paints in the center of the top and watch the paint spread from there. Paint near the center just spreads out to cover the rest of the top, but paint near the edges cascades down the sides. The homeomorphism on the sides/bottom just runs this picture backwards in time. Finally, repeat this on all concentric hypercubes. $\endgroup$ – Jason DeVito Nov 26 '19 at 16:34
  • $\begingroup$ Thanks a lot for your answer. I'm having some trouble at some point, maybe you can help me : when you say that for $x \in I^{n-1} \times \{-1\} $ but outside $[-1/2; 1/2]^{n-1} \times \{-1\}$, we define $f(x)= (h(x), \frac{d(x,g(x))}{d(x,h(x))})$ and you say that this part surjects onto $\partial I^{n-1} \times I$. Are you still assuming $I = [-1;1]$ ? Because I can't see how $\frac{d(x,g(x))}{d(x,h(x))}$ can take negative values. $\endgroup$ – Mathex Nov 27 '19 at 20:36
  • $\begingroup$ @Mathex: Oops. I guess at that point I'm thinking of $I$ as $[0,1]$. You can just change the second coordinate to $2\frac{d(x,g(x))}{d(x,h(x))} - 1$. That is, use a linear bijection $[0,1]\rightarrow [-1,1]$. Actually, looking back, the denominator should be a $d(g(x),h(x))$, not a $d(x,h(x))$. $\endgroup$ – Jason DeVito Nov 28 '19 at 2:04
  • $\begingroup$ Thanks a lot once gain for your help. I was on something else for a moment. Before I officially accept the answer, there is one thing I need to understand. When you say "the ray intersects $[-1/2; 1/2]\times \{-1\}$" you mean intersection with the boundary right ? Because I can see a whole segment of the ray that falls in this domain. Sorry to bother you with this but I want to make sure that I do understand every detail. Thanks ! $\endgroup$ – Mathex Dec 13 '19 at 23:59
  • $\begingroup$ One more thing : when you define $g(x)$ and $h(x)$ as the points where the ray from $p$ to $x$ intersects some boundary, you mean the actual points are of the form $(g(x), -1) \in \partial [-1/2;1/2]^{n-1} \times \{-1\}$ and $(h(x),-1) \in \partial I^{n-1} \times \{-1\}$, right ? $\endgroup$ – Mathex Dec 14 '19 at 1:19

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