Proposition 1

Let $X'$ be a normal topological space, let X be a subspace of $X'$, let A be a closed subspace of $X'$ contained in X, and let U be a neighbourhood of A in X.

Let us denote by $i_A$ the canonical injection of A into X$,i_U$ the canonical injection of U into X; let $j_A$ and $j_U$ denote the corresponding injections of Cyl($i_A$) and Cyl($i_U$) into $X\times \mathbf{I}$.

There exists a continuous mapping $r: X\times \mathbf{I}\rightarrow$ Cyl($i_U$) such that $j_U\circ$ $r(x) =x$ for every point $x\in j_A$(Cyl($i_A$)).

Let $U'$ be an open set of $X'$ such that $A\subset X\cap U'\subset U$. By definition of a normal space (TG, IX, p. 41), there exists an open neighbourhood $V'$ of A in $X'$ such that $A\subset V'\subset \overline{V'}\subset U'$ and a continuous function $\varphi ': X'\rightarrow \mathbf{I}$ which is equal to 1 at every point of A and to 0 at every point of $\complement V'$. Put $\varphi =\varphi '|X$ and $V = X\cap V'$. Let us also denote by $\alpha : U\times \mathbf{I}\rightarrow$ Cyl($i_U$) and $\beta : X\rightarrow$ Cyl($i_U$) the canonical mappings (III, p. 238).

Let $r$ be the mapping of $X\times \mathbf{I}$ into Cyl($i_U$) defined by

$\alpha (x,1-(1-t)\varphi (x))$ for $(x, t)\in U\times \mathbf{I}$,

$\beta (x)$ otherwise.

The mapping $r$ is continuous on $U\times \mathbf{I}$. For every point $(x, t)\in U\times \mathbf{I}$ such that $x\notin V$, we have $\varphi (x) = 0$, whence $r(x, t) =\alpha (x,1) =\beta (x)$. It follows that the mappings $r$ and $\beta \circ$ pr$_1$ coincide on $(X-V)\times \mathbf{I}$, which implies that $r$ is continuous on this subspace. Since U and the interior of X-V cover X, the mapping $r$ is continuous.

Let $y$ be a point of Cyl($i_A$) ; put $(x, t) =j_A(y)$. If $x\in$ A, $\varphi (x) = 1$ and $r(x, t) =\alpha (x, t)$, whence $j_U(r(x, t)) = (x, t)$. Otherwise, $t= 1$ and one has $r(x, t) =\alpha (x,1)$ if $x\in U$ and $r(x, t) =\beta (x)$ otherwise ; hence $j_U(r(x, t)) = (x, t)$ in this case. This entails that $j_U\circ r$ maps every point of $j_A$(Cyl($i_A$)) to itself, whence the proposition.