# Hatcher's Algebraic Topology 0.6(a) - Is this proof legit?

I am primarily not sure about my two-step homotopy construction:

Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the horizontal segment $[0, 1] \times \{0\}$ together with all the vertical segments $\{r\}\times[0, 1 - r]$ for $r$ a rational number in [0, 1]. Show that $X$ deformation retracts to any point in the segment $[0, 1]\times \{0\}$, but not to any other point. [See the preceding problem.]

Proof: Consider the homotopy $f_t = f_t^2 \circ f_t^1$ that $f_t^1: X \to [0,1] \times \{0\}$:

$$f_t^1(x,y) = (x,(1-t)y).$$

Hence we find the family of maps $f_t^1: X \to X, t \in I$, such that $f^1_0 = \mathbb{I}$ (the identity map), $f_1^1(X) = [0, 1]\times \{0\}$, and $f_t^1|[0, 1]\times \{0\}= \mathbb{I}$ for all $t$.

Now consider the homotopy $f_t^2: [0,1] \times \{0\} \to [p,0]$, where $p \in [0,1]$: $$f_t^2(x,0) = (x + t(p-x),0).$$

Hence we find the family of maps $f_t^2: [0,1] \times \{0\} \to [p,0], t \in I$, such that $f^2_0 = \mathbb{I}$, $f_1^2(X) = [p, 0]$, and $f_t^2|[p, 0]= \mathbb{I}$ for all $t$.

Now we show not to any other point. First reflect the fact that

Tube Lemma: Let $X$ and $Y$ be topological spaces with $Y$ compact, and consider the product space $X \times Y$. If $N$ is an open set containing a slice in $X \times Y$, then there exists a tube in $X \times Y$ containing this slice and contained in $N$.

And now we claim:

Lemma: Show that if a space $X$ deformation retracts to a point $x ∈ X$, then for each neighborhood $U$ of $x$ in $X$ $\exists$ a neighborhood $V ⊂ U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.

Proof: Given $X$ deformation retracts to a point $x \in X$, we know there is a family of maps $f_t: X \to X, t \in I$, such that $f_0 = \mathbb{I}$ (the identity map), $f_1(X) = c$, and $f_t|c= \mathbb{I}$ for all $t$.

To make sure $f_t(V) \subset V$, we need to find the appropriate $V$. By the definition of retraction, we know that $f_t(x) = x$, that is the slice $x \times I$ is contained in some open set $N$, and hence by tube lemma, $N$ also contains a tube containing the slice $x \times I$. Hence, let $V = N$.

Now we consider the inclusion map $i: V \to U$, and consider another family of maps: $\tilde{f}_t = f_t \circ i: X \to X, t \in I$ such that $\tilde{f}_0 = \mathbb{I} \circ i = i, \tilde{f}_1 = c \circ i$, $\tilde{f}_t|c= \mathbb{I}$ for all $t$.

Hence we constructed a homotopy map from inclusion map to a constant map, hence it is nullhomotopic.

Assume for contradiction that $X$ deformation retracts to some point $(x,y)$ that $x \in \mathbb{Q}$ and $y \in (0,1]$. Then by the previous question, for each neighborhood $U$ of $x$ in $X$ $\exists$ a neighborhood $V \subset U$ of $x$ such that the inclusion map $V \rightarrow U$ is nullhomotopic.

But we consider a small enough neiborhood $U$ of $x$ that does not contain points on $[0,1] \times \{0\}$. Then the neighborhood is not path-connected, and certainly can not homotopy to a constant map. Contradiction.