Let $X$ be a topological space and let $X'\subseteq X$ be a subspace of $X$. Let also $U$ be a path component $X$ such that.
I want to show that if $X'$ is a (strong) deformation retract of $X$ then $X'\cap U$ is a deformation retract of $U$.
My attempt is as follows: Let $$H:X\times[0,1]\to X$$ be the deformation retract of $X'$ in $X$, i.e.
- $\forall x\in X \ \ H(x,0)=x$,
- $\forall x\in X \ \ H(x,1)\in X'$, and
- $\forall x\in X'\forall t\in[0,1] \ \ H(x,t)=x$.
Now, I believe that since $U$ is a path-connected component of $X$, then if we restrict $H$ to $(X'\cap U)\times [0,1]$ we will take a deformation retract from $X'\cap U$ to $U$. That is because for each $u\in U$ the map $h_{u}:[0,1]\to X$ where $h_{u}(t)=H(u,t)$ is continuous and hence a path in $X$, which contains $h_{u}(0)=H(u,0)=u$. Therefore, since $U$ is a path-component of $X$, $h_{u}([0,1])\subseteq U$ and so $H(U\times[0,1])\subseteq U$.
I am not quite keen on with homotopy so I am not completely sure about all the implications of the above proof. Is there any flaw?