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Given $f : X \rightarrow Y$ I have the mapping cylinder $M_f = (X \times I) \sqcup Y / \sim$ where $I = [0,1]$ and $(x,0)_1$ is glued to $f(x)_2$ (I'd like to be pedantic for this verification so I'm adding subscripts resulting from the disjoint union). It is usually said that the mapping cylinder deformation retracts to the subspace $Y$; again being pedantic what must be meant is that $M_f$ deformation retracts to the subspace consisting of the points $\{y_2\}$ for $y \in Y-f(X)$ and $\{(x,0)_1,f(x)_2\}$ for $x \in X$, I'll call this subspace $C = \{\{y_2\}, \{(x,0)_1,f(x)_2\} \: | \: y \in Y - f(X), x \in X\}$.

My first question is that do we not have to be pedantic here because I assume that we could only loosely say $Y$ instead of this subspace if these two spaces were homeomorphic and for the obvious map between them to be a homeomoprhism in general we would require that $f$ be an injection?

Secondly, I'd like to verify that the expected deformation retraction is indeed one. $M_f$ consists of three types of points, $\{y_2\}$'s, $\{(x,s)_1\}$'s and $\{(x,0)_1,f(x)_2\}$'s; and the relevant map $M_f \times I \rightarrow M_f$ is $$\left(\{y_2\}, t\right) \mapsto \{y_2\} \:\:\: \forall y \in Y - f(X), t \in I,$$ $$\left(\{(x,0)_1,f(x)_2\},t\right) \mapsto \{(x,0)_1,f(x)_2\} \:\:\: \forall x \in X, t \in I,$$ $$\left(\{(x,s)_1\},t\right) \mapsto \{(x,(1-t)s)_1\} \:\:\: \forall x \in X, s \in (0,1], t \in [0,1),$$ $$\left(\{(x,s)_1\},1\right) \mapsto \{(x,0)_1,f(x)_2\} \:\:\: \forall x \in X, s \in (0,1].$$

That the map $M_f \rightarrow M_f$ retrieved by setting $t = 0$ is the identity map is clear, and so is that the map retrieved by setting $t = 1$ maps all of $M_f$ onto $C$ and finally that each map for a fixed $t$ restricted to $C$ is the identity map. The only concern is continuity. I was able to show that each map for a fixed $t$ is continuous (in a manner similar to how I attempted this problem), but not this map from $M_f \times I \rightarrow M_f$ that we have here. My attempt is as follows:

Let $U$ be any open subset of $M_f$; $U$ consists of points $\{y_2\}$ for $y$ in some $B \subseteq Y$, of points $\{(x,0)_1,f(x)_2\}$ for $x$ in some $A \subseteq X$ and of points $\{(x,s)_1\}$ for $(x,s)$ in some $Q \subseteq X \times (0,1]$. Using the definitions of the quotient and disjoint union topologies this tells us that $B \cup f(A)$ is open in $Y$ and that $A \times \{0\} \cup Q$ is open in $X \times I$. The preimage of $U$ under our map above is the union of the following

$$\{\{y_2\} \: | \: y \in B\} \times I,$$

(where the $\{y_2\}$'s come from)

$$\{\{(x,0)_1,f(x)_2\} \: | \: x \in A\} \times I,$$

$$\{\{(x,s)_1\} \: | \: x \in A, s \in (0,1]\} \times \{1\},$$

(where the $\{(x,0)_1,f(x)_2\}$'s come from), and

$$\{(\{(x,s)_1\},t) \: | \: (x,s,t) \in \varphi^{-1}[Q]\}$$

(where the $\{(x,s)_1\}$'s come from)

where $\varphi : X \times (0,1] \times [0,1) \rightarrow X \times I : (x,s,t) \mapsto (x,(1-t)s)$.

In the cases of the maps for each fixed $t$, the preimages worked out to be similar but simpler and openness wasn't too difficult to establish. Here all I have been able to say is that because $A \times \{0\} \cup Q$ is open in $X \times I$, its preimage under $\varphi$, which is in fact just $\varphi^{-1}[Q]$ because the second coordinate of any image point cannot be zero, is open in $X \times (0,1] \times [0,1)$ and as a result in $X \times [0,1] \times [0,1]$, this then implies that our final set $\{(\{(x,s)_1\},t) \: | \: (x,s,t) \in \varphi^{-1}[Q]\}$ is open in $M_f \times I$. From here however I haven't been able to get further. I've tried picking points in one of the four sets and then trying to combine subsets of the four sets in various ways to find an open set containing the point within our four sets and I've noted some small things like the first two of our four sets may be considered together in the form $E \cup I$ with $E \subseteq M_f$ and the pullback across the quotient projection map and then the disjoint union inclusion map of $E$ gives $B \cup f(A)$ in $Y$ which is known to be open but we get $A \times \{0\}$ in $X \times I$.

I decided to do this verification, and in this pedantic a manner, for practice. I'd very much like to move on but it's bugging me and I'd really appreciate some help.

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    $\begingroup$ Not a complete answer, but a few points that might help simplify your task. First, given any quotient space $E/{\sim}$, you have $E/{\sim}\times I \cong (E \times I)/{\sim}$, where the second $\sim$ is, by abuse of notation, the obvious equivalence relation on $E \times I$. (Local) compactness of $I$ is essential here. Second, wherever possible, use the fact that cont. maps from a quotient space $E/{\sim}$ to a space $F$ correspond precisely to cont. maps $E \to F$ that respect the equivalence relation $\sim$. Third, a cont. map $E \sqcup F \to G$ is the same as maps $E \to G$ and $F \to G$. $\endgroup$
    – user120974
    Jan 13 '14 at 12:06
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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 $I$.

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  • $\begingroup$ Sorry I just realised I completely forgot to accept your answer and thank you! This is much delayed but your answered helped me very much, so again thank you $\endgroup$
    – user34832
    May 11 '14 at 11:27
  • $\begingroup$ “So it induces a continuous map H”. What the hell are you talking about? You never define what H or G is! $\endgroup$ Jul 24 '20 at 2:26

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