Find a space \(X\) and a point \(x_0\) of \(X\) such that the inclusion \(\ts{x_0} \to X\) is a homotopy equivalence, but \(\ts{x_0}\) is not a deformation retract of \(X\). [Hint: Let \(X\) be the subspace of \(\mathbb{R}^2\) that is the union of the line segments \((1/n) \times I\), for \(n \in \mathbb{Z}_+\), the line segment \(0 \times I\), and the line segment \(I \times 0\); let \(x_0\) be the point \((0, 1)\). If \(\ts{x_0}\) is a deformation retract of \(X\), show that for any neighborhood \(U\) of \(x_0\), the path component of \(U\) containing \(x_0\) contains a neighborhood of \(x_0\).]
Solution
Let \(X\subset\mathbb R^2\) be the comb space from the hint: \[ X=(I\times\{0\})\cup(\{0\}\times I)\cup\bigcup_{n\ge1}(\{1/n\}\times I), \] and let \(x_0=(0,1)\).
The space \(X\) is contractible. One explicit contraction first moves every point vertically down to the base and then moves along the base to \((0,0)\); concatenating this with the path in \(\{0\}\times I\) from \((0,0)\) to \(x_0\) gives a homotopy of the identity to the constant map at \(x_0\). Hence the inclusion \(\{x_0\}\hookrightarrow X\) is a homotopy equivalence.
We show that \(\{x_0\}\) is not a deformation retract. Suppose \(H:X\times I\to X\) were a deformation retraction onto \(x_0\), so \(H(x,0)=x\), \(H(x,1)=x_0\), and \(H(x_0,t)=x_0\).
A standard consequence of continuity at the compact set \(\{x_0\}\times I\) is the following: for every neighborhood \(U\) of \(x_0\), there is a neighborhood \(V\) of \(x_0\) such that \[ H(V\times I)\subset U. \] Indeed, cover \(\{x_0\}\times I\) by finitely many product neighborhoods on which \(H\) lands in \(U\), then intersect their \(X\)-factors. For each \(v\in V\), the path \(t\mapsto H(v,t)\) lies in \(U\) and joins \(v\) to \(x_0\). Thus the path component of \(U\) containing \(x_0\) contains the neighborhood \(V\) of \(x_0\).
Now take \[ U=X\cap\{(x,y):y>1/2\}. \] The path component of \(U\) containing \(x_0\) is exactly \(\{0\}\times(1/2,1]\): every other vertical tooth \(\{1/n\}\times(1/2,1]\) is a distinct path component because the horizontal base \(y=0\) has been removed. This component contains no neighborhood of \(x_0\) in \(X\), since every neighborhood of \((0,1)\) meets infinitely many teeth \(x=1/n\). Contradiction. Hence \(\{x_0\}\) is not a deformation retract.