Proving continuity of homotopy of clutching functions

Question

This is in page 45 of Hatcher's, that there is a linear homotopy between clutchig functions .

I could not workout/justify the linear homotopy. Is there an easy way to see the continuity of homotopy?

EDIT: I have found on page 63 of this homotopy. But I do not follow at all the continuity. How did we even assign a norm?

Answer 1

Hatcher considers the set $\text{End}(E \times S^1)$ of all bundle endomorphism on $E \times S^1$ and endows it with a norm $\lVert - \rVert$. He then states that the subspace $\text{Aut}(E \times S^1)$ of all bundle automorphism on $E \times S^1$ is open in this normed linear space. Given any two $f_0, f_1 \in \text{End}(E \times S^1)$, the linear path $u : I \to \text{End}(E \times S^1), u(t) = (1-t)f_0 + t f_1,$ is continuous. If $f_0,f_1$ belong to $\text{Aut}(E \times S^1)$, then $u$ may leave $\text{Aut}(E \times S^1)$, but if they are sufficiently close (which means that they are contained in a sufficiently small open $\lVert - \rVert$-ball), then $u$ will stay in $\text{Aut}(E \times S^1)$. Taking the adjoint of $u$ provides a function $u^* :E \times S^1 \times I \to E \times S^1$ which is the desired homotopy provided it is continuous. In fact Hatcher has a gap at this point: He doesn't prove the continuity of $u^*$.

We can regard $\text{End}(E \times S^1)$ as a subset of the function space $(E \times S^1)^{E \times S^1}$ endowed with the compact open-topology. If we would know that the compact-open topology on $\text{End}(E \times S^1)$ is coarser than the norm-topology, then also $u : I \to \text{End}(E \times S^1) \hookrightarrow (E \times S^1)^{E \times S^1}$ would be continuous and we could conclude that $u^*$ is continuous because $E \times S^1$ is locally compact (note that $X$ is assumed to be compact).

I am quite optimistic that this can be shown, and I believe the norm topology even agrees with the compact-open topology. However, I shall not further pursue this.

Answer 2

This is a solution I came up. I am happy for any criticism.

(I) Some properties of transition functions:

We observe $p \times 1: E \times S^1 \rightarrow X \times S^1$ is a vector bundle. Locally, there are open subsets $U \subseteq X, V \subseteq S^1$, and homeomorphism charts, $$h : p^{-1}(U) \times V \rightarrow (U \times V) \times \mathbb C^n .$$

A clutching function induces a collection of transition function: For simplicity, let $U':=U \times V$. So the map $$hlh^{-1}:U' \times \mathbb C^n \rightarrow U' \times \mathbb C^n, (x,v) \mapsto (x,A_l(x)v)$$ is a continuous function, where $A_l: U' \rightarrow GL_n(\mathbb C)$ is a continuous map. This can be seen by choosing a basis of $\mathbb C$, and note that each component of $A_l$ is given by the composition $$ (u,e_j) \mapsto (u,A_l(u)e_j) \mapsto \pi_i(A_l(u)e_j), \pi_i:\mathbb C^n \rightarrow \mathbb C$$

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(II) Constructing homotopy Now for defining the homotopy $E \times S^1 \times I \rightarrow E \times S$. We may explicitly define the map as $$ (x,t) \mapsto l(x)(1-t)+tf(x).$$ To show continuity of this map we break it down locally. Fix a cover of base space $X$, $\{U_i \times S^1 \}$. Fix some $U$ in the cover. Then we have transition functions $A_l, A_f:U \rightarrow GL_n(\mathbb C)$. we first define $$g: U \times S^1 \times I \times \mathbb C^n \rightarrow U \times S^1 \times \mathbb C^n, (x,z,v) \mapsto (x,z, (A_l(x,z)(1-t)+tA_f(x,z))v) $$

This map is continuous as we argued in (I), noting that component why it is a polynomial function of $t$ and components of matrix (which we have previously shown is continuous), hence continuous.

Hence our map can be broken as follows, $$ \pi^{-1}(U) \times S^1 \times I \xrightarrow{h} U \times S^1 \times I \times \mathbb {C^n} \xrightarrow{g} U \times S^I \times \mathbb C^n \xrightarrow{h^{-1}} \pi^{-1}(U) \times S^1.$$

which is continuous.

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(III) A summary Essentially the proof uses locality - and that we can transport the structure to two data:

1. a transition function $U \rightarrow GL_n(\mathbb C)$, which is continuous.

2. A map between vector space $U \times \mathbb C^n \rightarrow U \times \mathbb C^n$.

2'. When chosen with basis, the components of transition matrix is continuous.

Answer 3

Here is one more answer. Let $\xi = (E,p,X), \eta = (F,q,Y)$ be vector bundles and $(f_i,\varphi) : \xi \to \eta$ be two bundle maps over the same $\varphi : X \to Y$ (see e.g. Husemoller's book; we have $q f_i = \varphi p$ and for each $x \in X$ the map $f_{i,x}$ from the fiber over $x$ to the fiber over $\varphi(x)$ is linear).

Then the fiberwise linear homotopy $H : E \times I \to F, H(v_x,t) = (1-t)f_0(v_x) + tf_1(v_x)$ , where $v_x \in p^{-1}(x)$, is continuous. That is, all bundle maps over the same map on the base spaces are homotopic.

This is obvious for trivial bundles. In the general case we can use local trivializations to see that $H$ is locally continuous and hence continuous. Note that under local trivializations fiberwise linear homotopies are transformed locally to fiberwise linear homotopies between trivial bundles.

Hatcher applies this to the case of bundle maps over $id : X \times S^1 \to X \times S^1$. The interesting point is not that all elements of $\text{End}(E \times S^1)$ are homotopic, but that fiberwise linear homotopies stay in $\text{Aut}(E \times S^1)$ provided $f_0,f_1$ are sufficiently close automorphisms. This is why he introduced the norm on $\text{End}(E \times S^1)$. In fact, he uses the convexity of norm-balls to assure that the linear homotopy between $f_0$ and $f_1$ stays in $\text{Aut}(E \times S^1)$.