I was trying to prove some theorem in topology, which I am stuck on for quite some time. I have though of some result that seems very intuitive and would be very useful, but I can not seem to justify it rigorously. I would be very happy if I could get some help with it.
Suppose $X$ is a perfectly normal space (not sure if all of that is needed). Next, suppose we have a closed subset $A \subset X$ and a strong deformation retraction $H: X \times I \to X$ of $X$ to $A$. Further, suppose we have two open neighbourhood $B$ and $C$ of $\text{Int}(A)$, such that $\text{Int}(A) \subseteq B \subseteq C$. The sets $B$ and $C$ have an additional property, that their boundaries only intersect in $A$. Can we create a continuous map $F: X \times I \to X$, for which $F|_{X \times 0} = \text{id}$, $F|_{B} = H|_{B}$ and $F|_{(X-C) \times 1}$ is identity, so that it performs the deformation retraction $H$ fully on $B$, but not at all outside of $C$.
I have been trying to prove this for quite some time and it seems very natural, but I couldn't do it. If $B$ and $C$ shared no common boundary, it would be pretty easy; we could just define a function which is $1$ on the closure of $B$ and $0$ outside of $C$, then make a convex combination of $H$.
If it is not correct, maybe some additional assumptions would help, which might be present in my situation, so please provide a counterexample.
You actually can get away with assuming only that $X$ is completely normal, i.e., hereditarily normal (this is weaker than perfect normality).
The reason is you can then invoke normality of the subspace $Y=X\backslash (\partial B\cap \partial C)$. Note that $\overline{B}\cap Y$ and $\overline{X\backslash C}\cap Y$ are disjoint and closed relative to $Y$, so by normality (and Urysohn's lemma) there is some continuous function $f\colon Y\to I$ with $f|_{\overline{B}\cap Y}\equiv 1$ and $f|_{\overline{X\backslash C}\cap Y}\equiv 0$.
Define $F\colon X\times I\to X$ by $$ F(x,t)=\begin{cases} H(x,f(x)t) & x\in Y\text{,}\\ x & x \in \partial B\cap \partial C\text{.} \end{cases} $$
Then by inspection $F$ has all of the desired properties, with only continuity remaining to be proven.
To see that $F$ is continuous, we observe that since $Y\times I$ is open, $F$ is continuous at each point in $Y\times I$. On the other hand, if $(x,t)\in (\partial B\cap \partial C)\times I$, then let $U$ be a neighborhood of $F(x,t)=x$. Since $H(x,t)=x$ for each $t\in I$, by continuity of $H$, $H^{-1}(U)$ is a neighborhood of $\{x\}\times I$. By compactness of $I$, $H^{-1}(U)$ contains a rectangular neighborhood of the form $V\times I$ with $x\in V$.
But for $(z,s)\in (V\cap U)\times I$, we either have $F(z,s)=z\in V\cap U\subseteq U$, or $F(z,s)=H(z,f(z)s)\in H(V\times I)\subseteq U$. Thus we see that $$(x,t)\in (V\cap U)\times I\subseteq F^{-1}(U),$$ and so the preimage of every neighborhood of $F(x,t)$ is a neighborhood of $(x,t)$, hence $F$ is continuous at $(x,t)$.