Hatcher, *Algebraic Topology*, Chapter 0

**5. Show that if a space deformation retracts to a point , then for each neighborhood of in there exists a neighborhood of such that the inclusion map is nullhomotopic.**^{1}

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

Without loss of generality, suppose the family of maps for which represents the deformation retract of onto . In particular, then, , , and for all and so it follows that is the constant map and that fixes for all . Clearly, a deformation in this (strong) sense implies a deformation retract in the weak sense, so by previous results, there exists an inclusion for which and . Let be the homotopy connecting to so that and for all .

Because is a deformation retract of onto the point , the family can be thought of as a *shrinking* of onto the subset , and because the family is continuous in both and (by definition of homotopy), it follows that there exists an element for which for and for . Let . It suffices to show that the inclusion is homotopic to a constant map.

To that end, write , , and define so that

for all ^{2}.

In particular, then, as . Moreover, for all , and because , . Finally, because is a (strong) deformation retract, because is defined in terms of , and because is defined in terms of , it’s easily confirmed that for all and hence that is a *weak* deformation retract. Therefore, from the same problem cited above (see here), there exists an inclusion map for which . In particular, is null-homotopic.

At this point, all the ingredients are in place. The fact that the family homotopes to implies that restricted to homotopes to . Recall, however, that is the constant map , and so . Thus, . Combining this with the fact that yields that . Tracing through the chaos, the fact that is easily recovered, whereby implies that the inclusion is, indeed, homotopic to the constant map collapsing the entire domain to . Hence, the result holds.

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 “” is less-notationally-mind-numbing way of defining .

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!

$H^{-1}(U)$ 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.