Proposition 5

The mapping $\alpha_f$ induces a homeomorphism of $X\times [0,1[$ onto an open subset of Cyl($f$). The mapping $\beta_f$ defines a homeomorphism of Y onto the complement of this open subset. The mapping $\beta_f\circ \gamma_f$ is a continuous retraction of Cyl($f$) onto $\beta_f(Y)$.

The set $X\times [0,1[$ is an open subset of U, saturated for the relation R, and the equivalence relation induced by R in $X\times [0,1[$ is the equality relation. The first assertion follows from this (TG, I, p. 23, cor. 10).

The complement of $\alpha_f(X\times [0,1[)$ is $\beta_f(Y)$. Since one has $\gamma_f\circ \beta_f=$ Id$_Y,\beta_f$ defines a homeomorphism of Y onto $\beta_f(Y)$ and $\beta_f\circ \gamma_f$ is a continuous retraction of the canonical injection of $\beta_f(Y)$ into Cyl($f$).

The closed subspace $\beta_f(Y)$ of Cyl($f$) is called the basis of the cylinder of $f$. The mapping $\beta_f\circ \gamma_f$ is called the canonical retraction of Cyl($f$) onto its basis.

Consider the mappings $\sigma_1: U_1\times \mathbf{I}\rightarrow$ Cyl($f$) and $\sigma_2: U_2\times \mathbf{I}\rightarrow$ Cyl($f$) defined by

$\sigma_1((x, s), t) =\alpha_f(x,(1-t)s+t)$ for $(x, s)\in X\times \mathbf{I}$ and $t\in \mathbf{I}$

$\sigma_2(y, t) =\beta_f(y)$ for $y\in Y$ and $t\in \mathbf{I}$.

They are continuous. For $x\in X$ and $t\in \mathbf{I}$, one has

There therefore exists a unique mapping $\sigma_f$ of Cyl($f$)$\times \mathbf{I}$ into Cyl($f$) such that $\sigma_f\circ (\alpha_f\times$ Id$_{\mathbf{I}}) =\sigma_1$ and $\sigma_f\circ (\beta_f\times$ Id$_{\mathbf{I}}) =\sigma_2$. The mapping $\sigma_f$ is continuous (I, p. 19, prop. 10).