Remark 3

If the spaces X and Y are separated, the same is true of the mapping cylinder of the mapping $f$. Indeed, let $z$ and $z'$ be two distinct points of Cyl($f$) and let us prove that they possess disjoint neighbourhoods. Distinguish three cases.

– There exist $(x, t)$ and $(x', t')\in X\times [0,1[$ such that $z=\alpha_f(x, t)$ and $z'=\alpha_f(x', t')$.

In this case, the assertion follows from the fact that the space $X\times [0,1[$ is separated (TG, I, p. 54, prop. 7) and that the mapping $\alpha_f$ induces a homeomorphism of this space onto an open subspace of Cyl($f$).

– There exists $(x, t)\in X\times [0,1[$ such that $z=\alpha_f(x, t)$ and $y'\in Y$ such that $z'=\beta_f(y')$.

Then, $\alpha_f(X\times [0,\frac{t+1}{2}[)$ and Cyl($f$)$-\alpha_f(X\times [0,\frac{t+1}{2}])$ are disjoint open neighbourhoods of $z$ and $z'$ in Cyl($f$).

– There exist $y$ and $y'\in Y$ such that $z=\beta_f(y),z'=\beta_f(y')$.

In this case, $y=\not y'$; since Y is separated, there exists an open neighbourhood V of $y$ in Y and an open neighbourhood $V'$ of $y'$ in Y such that $V\cap V'=\emptyset$. Then, $(\beta_f\circ \gamma_f)^{-1}(V)$ and $(\beta_f\circ \gamma_f)^{-1}(V')$ are disjoint open subsets of Cyl($f$) containing respectively $y$ and $y'$.