Suppose that a space $X$ deformation retracts to a point $x_0\in X$. I want to prove that for any open neighbourhood $U$ of $x_0$, there exists an open neighbourhood $V$ of $x_0$ with $V\subset U$, such that the inclusion of $V$ in $U$ is based homotopic to the constant map. Any ideas?
Deformation retracts to a point implies inclusion on open neighbourhoods is locally based homotopic to constant map.
algebraic-topology
Best Answer
Assume that $F:X\times I\rightarrow X$ is a contraction of a space $X$ relative to a point $\ast\in X$. i.e. $F$ is a continuous map with $F(x,0)=x$ and $F(x,1)=\ast$ for all $x\in X$ and $F(\ast,t)=\ast$ for all $t\in I$.
Proof: Since $F(\ast\times I)=\ast\in U$ we have $\ast\times I\subseteq F^{-1}(U)$. Note that $F^{-1}(U)$ is open since $F$ is continuous. Using the compactness of $I$ we appeal to the tube lemma to find an open set $V\subset X$ with $\ast\times I\subseteq V\times I\subseteq F^{-1}(U)$. Then $F(V\times I)\subseteq U$, so $F|_{V\times I}$ is the claimed contraction of $U$ in $V$ relative to $\ast$.