I have $X$ retract by retraction $r$ of its open neighboorhood $Y\in\mathbb{R}^n$.
I'd like to write that, if $O\subset X$ is an open set, so $r^{-1}(O)$ is an open set in $\mathbb{R}^n$. But $r$ continuous only means that $r^{-1}(O)$ is an open set in $Y$. However, I feel that in this case the topology of $Y$ is the same that the one usual in $\mathbb{R}^n$.
Could someone give me a light on this?
Thank you so much.
The setup is that there is an embedding $j:X\hookrightarrow\mathbb{R}^n$, an open set $Y\subset\mathbb{R}^n$ with $X\cong j(X)\subseteq Y$, and a continuous map $r:Y\rightarrow X$. The question regards the topology on $Y$. Suppose that $V\subset Y$ is any subset which is open in $Y$. Because $Y$ has the subspace topology inherited from $\mathbb{R}^n$, this means that there is an open set $\widetilde V\subset\mathbb{R}^n$ such that $V=\widetilde V\cap Y$. But $\widetilde V$ and $Y$ are both open in $\mathbb{R}^n$, and hence so is their intersection $V$. In particular we can apply this when $V=r^{-1}(U)$ with $U\subseteq X$ open.