For $X$ a topological space and $A$ a subspace, we say that $(X,A)$ is a good pair if $A$ is closed and there is a neighborhood of $A$ which deformation retracts to $A$. Let $X$ be a topological space, and $B \subseteq A \subseteq X$ be two subspaces. If $(X,A)$ and $(A,B)$ are good pairs, is $(X,B)$ a good pair?
My guess is that it is not true in general. We can define a straight deformation retract as a deformation retract $f: U \times I \to U$ such that $f(f(u,t),1) = f(u,1)$ for any $u \in U$, any $t \in I$. Let $U$ be the neighborhood of $A$, with the deformation retract $f : U \times I \to U$, and $V$ the neighbordhood of $B$ in $A$. A counterexample, if it exists, should be weird, because if $f$ is a straight deformation retract, and if $V$ is open, then $f^{-1}(V)$ is an open neighborhood which deformation retracts to $V$, hence which deformation retracts to $B$.
