Hatcher 0.5

Hatcher, Algebraic Topology, Chapter 0

5. Show that if a space X deformation retracts to a point x\in X, then for each neighborhood U of x in X there exists a neighborhood V\subset U of x such that the inclusion map V\hookrightarrow U is nullhomotopic.1

Proof. Let X be a topological space, let x be an element of X onto which X deformation retracts, and let U\subset X be an arbitrary neighborhood of x in X. As per the statement of the problem, it suffices to find a neighborhood V\subset U of x for which the inclusion map \iota:V\to U is nullhomotopic, meaning that there exists a homotopy connecting \iota with a constant map.

Without loss of generality, suppose the family of maps F:X\times I\to X for which F(x,t)=f_t(x) represents the deformation retract of X onto x\in X. In particular, then, f_0=id_X, f_1(X)=x, and f_t(x)=x for all t\in I and so it follows that f_1 is the constant map X\to x and that f_t fixes x for all t\in I. Clearly, a deformation in this (strong) sense implies a deformation retract in the weak sense, so by previous results, there exists an inclusion \iota:\{x\}\to X for which f_1\iota\simeq 1_{\{x\}} and \iota f_1\simeq 1_X. Let G:X\times I\to X be the homotopy connecting \iota f_1 to 1_X so that G(x,0)=g_0(x)=1_X(x)=x and G(x,1)=g_1(x)=(\iota\circ f_1)(x) for all x\in X.

Because \{f_t\}_{t\in I} is a deformation retract of X onto the point x\in X, the family \{f_t\} can be thought of as a shrinking of X onto the subset \{x\}, and because the family \{f_t\} is continuous in both t and x (by definition of homotopy), it follows that there exists an element T\in I for which g_t^{-1}(U)\subsetneq U for t\geq T and g_t^{-1}(U)=U for t<T. Let V=g_T^{-1}(U). It suffices to show that the inclusion V\hookrightarrow U is homotopic to a constant map.

To that end, write t=kT, k\in[0,1], and define H:U\times I\to V so that

H(u,k)=h_k(u)=g_{kT}(x) for all x\in X\cap U2.

In particular, then, h_0(u)=g_0(u)=u as g_0=id_X. Moreover, h_1(u)=g_T(u) for all u\in U, and because V=g_T^{-1}(U), h_1(U)\subset V. Finally, because f_t is a (strong) deformation retract, because G is defined in terms of f_t, and because H is defined in terms of G, it’s easily confirmed that h_t(V)\subset V for all t and hence that h_t is a weak deformation retract. Therefore, from the same problem cited above (see here), there exists an inclusion map j:\{x\}\to V for which h_1\circ j\simeq 1_{\{x\}}. In particular, h_1\circ j is null-homotopic.

At this point, all the ingredients are in place. The fact that the family \{f_t\} homotopes 1_X\simeq \iota\circ f_1 to 1_{\{x\}} implies that \{f_t\} restricted to V\subset U homotopes 1_V\simeq(\iota\circ f_1)|_V to 1_{\{x\}}. Recall, however, that f_1 is the constant map X\to\{x\}, and so (\iota\circ f_1)(V)=\iota(x). Thus, \iota\simeq 1_V\simeq 1_{\{x\}}. Combining this with the fact that h_1\circ j\simeq 1_{\{x\}} yields that h_1\circ j\simeq \iota. Tracing through the chaos, the fact that h_1=g_T is easily recovered, whereby h_1\circ j\simeq \iota implies that the inclusion V\hookrightarrow U is, indeed, homotopic to the constant map collapsing the entire domain to x. Hence, the result holds.   \square

    1. Ideas for this proof were borrowed from solutions posted by Tarun Chitra to his personal webpage, http://www.tarunchitra.com. This particular problem was posted as part of the document: http://tarunchitra.com/papers/6510/hw5.pdf

    2. The restriction “x\in X\cap U” is less-notationally-mind-numbing way of defining h_t(u)=g_t|_U(x).

2 thoughts on “Hatcher 0.5

  1. I feel like this proof can be simplified dramatically. As mentioned above, I used Tarun’s ideas, which means I in turn used a lot of his perspectives/procedures. Given the ideas, however, I feel like I can make this proof more “my own” with some time, and in doing so, I’m imagining a result that’s somewhat simpler. Stay tuned!

  2. $H^{-1}(U)$ is a neighborhood of $\{x\} \times [0,1]$, so by compacity, it contains some cylinder $V \times [0,1]$ where $V$ is a neighborhood of $x$. For each $y \in V$, we have that $H(y,t) \in U$ for all $t$, so $H$ restricts to a homotopy : $V \times [0,1] \rightarrow U$ that verifies what we asked.

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