Verifying continuity of the deformation retraction of the mapping cylinder

Solution 1:

First note that we can indeed regard $Y$ as a subspace of $M_f$. This is because the map $\bar i:Y\to M_f$ which is the composition $Y\hookrightarrow X\times I\sqcup Y\xrightarrow q M_f$ is

• injective: Assume that $y_1,y_2$ are two points in $Y$. Then each $y_k$ is identified with all the points $(x,0)\in X\times I$ such that $f(x)=y_k$. The only way that $y_1$ and $y_2$ could be identified is by means of a finite sequence of identification "$(x,0)\sim y$ whenever $f(x)=y$". Induction over the number of such basic identifications shows that all $y$ which appear in this sequence must denote the same element in $Y$. So $\bar i(y_1)=\bar i(y_2)$ only if $y_1=y_2$.
• closed: Take a closed $F\subseteq Y$. Then its saturation is the set $\hat F=F\sqcup f^{-1}(F)\times\{0\}$, which is closed in $X\times I\sqcup Y$. Thus its image $q(\hat F)$ is closed in $M_f$ and $q(\hat F)\cap q(Y)=q(F)=\bar i(F)$ is thus a closed subset of $\bar i(Y)$

That means that $\bar i:Y\to\bar i(Y)$ is a homeomorphism. For convenience we talk about a point $y\in M_f$ when we actually mean the class $\{y,(x,0)\mid f(x)=y\}$.

Now you want a homotopy $H:M_f\times I\to M_f$. Put $h_t=H(-,t)$. Since $M_f$ is a quotient space, each $h_t:M_f\to M_f$ is determined by a map $g_t:Z:=Y\sqcup X\times I\to M_f$ which respects the relation. Now define $\sim'$ to be the relation on $Z\times I$ such that $(a,t)\sim'(a',t')\iff a\sim a'\wedge t=t'$. We see that if $G(a,t)=g_t(a)$ is continuous on $Z\times I$, then it respects $\sim'$ iff each $g_t$ respects $\sim$. So it induces a continuous map $H:Z\times I/\sim'\to M_f$. However, $(Z\times I)/\sim'$ is not the same as $(Z/\sim)\times I$. We merely have a continuous bijection $(Z\times I)/\sim'\to(Z/\sim)\times I$ induced by $q\times\text{id}:Z\times I\to Z/\sim\times I$, and there is no reason why it should be open. However, in this situation it is, and this is due to the local compactness of $I$.