# Homotopy equivalence of mapping cylinders

I was working on a problem in Hatcher´s Algebraic Topology. I could not entirely solve it, but I was able to reduce the rest of the problem to the following:

Let $$X$$ and $$Y$$ be topological spaces and let $$f:X\to Y$$ be a continuous map between them. Moreover, let $$h:X\to X$$ be a continuous map homotopic to $$\text{id}_X$$. Consider the mapping cylinders $$M_f$$ and $$M_{f\circ h}$$ and the map $$$$M_{f\circ h}\to M_f$$$$ given by \begin{align*} X\times I\ni (x,t)&\mapsto (h(x),t) \in X\times I \\ Y\ni y &\mapsto y\in Y \end{align*} I want to show that this map is a homotopy equivalence.

I would be glad for references, proofs or (if it is not correct) even examples showing that this is not true in general.

• Since $h$ is a homotopy equivalence, there is some other map $g: X \to X$ which has that $gh \simeq \text{id}_X$ and $hg\simeq \text{id}_X$. Then you should define a map $M_f \to M_{fh}$ sending $(x,t) \mapsto (g(x),t)$. This will be the desired homotopy inverse to the map you described. Commented Aug 30, 2019 at 20:43
• @desiigner What you said does not work. As $h$ is homotopic to $id_X$ you can choose $g=id_X$. But then the map you wrote down is not well defined, as in $M_f$ $(x,0)=f(x)$ but in $M_{f\circ h}$ $(g(x),0)=(x,0)$ is not $f(x)$. Commented Aug 30, 2019 at 21:19

$$\require{AMScd}$$

Begin by choosing a homotopy $$H:id_X\Rightarrow h.$$ Then the diagram $$\begin{CD} Y @id_X>>X\\ @Vid_YVV\stackrel{fH}{\Rightarrow} @VVid_XV@VVid_XV \\ Y @<{fh}<< X@>>id_X>X \end{CD}$$ induces a map $$\theta:M_f\rightarrow M_{fh}$$. I don't think the notation above is in Hatcher's book, but it's common to many of us. It is used to describe maps between double mapping cylinders, or homotopy pushouts, the first of which Hatcher seems to discuss briefly in $$\S$$ 4.G Gluing Constructions, around pg. 456.

To explain the diagram consider the top row $$Y\xleftarrow{f} X\xrightarrow{id_X} X$$. This means take the disjoint union of the space $$Y$$ on the left, the space $$X$$ on the right, and the cylinder $$X\times I$$ on the space in the middle. Then identify $$f(x)\in Y$$ with $$(x,0)\in X\times I$$ and $$x\in X$$ with $$(x,1)\in X\times I$$. This rather special case in which one map is the identity just gives you up to homeomorphism the standard mapping cylinder $$M_{f}=\left(X\times I\bigsqcup Y\right)\bigg/[X\times I\ni (x,0)\sim f(x)\in Y].$$ One thing to be careful with, however, is the orientations of your cylinders and homotopies. Similarly form $$M_{fh}$$ from the bottom row of the diagram.

Now to interpret the map that the diagram describes is simple. The left-hand vertical map describes what to do on $$Y$$. The prescribed homotopy $$fH$$ filling in the left-hand square tells you what to do on the bottom of the cylinder. The vertical map in the middle is the identity and says do nothing in the middle of the cylinder. The homotopy in the right-hand square is the trivial homotopy and tells you to do nothing on the top of the cylinder (apart from perhaps reparametrisation).

Hence the diagram describes the map $$\theta:M_f\rightarrow M_{fh}$$ given by $$\theta(y)=y,\qquad \theta(x,t)=\begin{cases}f(H(x,2t))\in Y&0\leq t\leq \frac{1}{2}\\(x,2t-1)\in X\times I&\frac{1}{2}\leq t\leq 1.\end{cases}$$

To get a map in the other direction we have several choices. For example we can use the diagram $$\begin{CD} Y @id_X>>X\\ @Vid_YVV @VhVV\stackrel{-H}{\Rightarrow} @VVid_XV \\ Y @<{f}<< X@>>id_X>X \end{CD}$$ which describes a map which does nothing on the bottom of the cylinder, is $$h\times 1$$ on the middle and is the inverse homotopy $$-H:h\Rightarrow id_X$$ on the top. That is, it is the map $$y\mapsto y,\qquad (x,t)\mapsto\begin{cases}(h(x),2t)&0\leq t\leq \frac{1}{2}\\(H(x,2-2t),1)&\frac{1}{2}\leq t\leq 1.\end{cases}$$ This map was my original (pre-edit) choice. However, not every map is going to be as easy to work with on a point-set level, and the utility of the diagramatic construction is that by 'sliding' the homotopy $$-H$$ in the right-hand square into the left-hand square, where it becomes the homotopy $$-fH$$, we get a linear homotopy of the previous map to the one described by the diagram $$\begin{CD} Y @id_X>>X\\ @Vid_YVV\stackrel{-fH}{\Rightarrow} @VVid_XV @VVid_XV \\ Y @<{f}<< X@>>id_X>X. \end{CD}$$ This is the diagram we will use to define the inverse: $$\varphi:M_{fh}\rightarrow M_f,\qquad y\mapsto y,\qquad (x,t)\mapsto\begin{cases}fH(x,1-2t)&0\leq t\leq\frac{1}{2}\\(x,2t-1)&\frac{1}{2}\leq t\leq 1\end{cases}$$ (A homotopy $$G$$ between these two maps is indeed easy to write down, and, as promised, corresponds simply to sliding $$-H$$ down the cylinder, $$G_s(x,t)=\begin{cases}(h(x),2t)&0\leq t\leq \frac{1-s}{2}\\(H(x,2-s-2t),1-s)&\frac{1-s}{2}\leq t\leq\frac{2-s}{2}\\(x,2t-1)&\frac{2-s}{2}\leq t\leq 1.)\end{cases}$$ Now we'll show that with our (new) choice of $$\varphi$$, the two maps $$\varphi,\theta$$ are inverse homotopy equivalences. To find out what the maps compose to we can work explicitly, or just paste the defining diagrams for each map on top of each other. For example the map $$\theta\varphi:M_{fh}\rightarrow M_{fh}$$ is generated by the diagram $$\begin{CD} Y @id_X>>X\\ @Vid_YVV\stackrel{-fH+fH}{\Rightarrow} @VVid_XV@VVid_XV \\ Y @<{fh}<< X@>>id_X>X \end{CD}$$ so that $$\theta\varphi(y)=\theta(y)=y,\qquad\theta\varphi(x,t)=\begin{cases}fH(x,1-3t)\in Y&0\leq t\leq \frac{1}{3}\\fH(x,3t-1)\in Y&\frac{1}{3}\leq t\leq \frac{2}{3}\\ (x,3t-2)\in X\times I&\frac{2}{3}\leq t\leq 1.\end{cases}$$ Now technically this isn't exactly the indicated composite, but rather something very easily seen to be homotopic to it (it just arranges the intervals differently by linear homotopies).

Now this composite is clearly homotopic to the identity on $$M_{fh}$$ by matching up the ends where the two $$\pm fH(x,-)$$ meet. Indeed, we have the map $$F_\bullet:M_{fh}\times I\rightarrow M_{fh}$$, which is the identity on $$Y$$, $$F_s(y)=y$$, $$\forall y\in Y,s\in I$$, and on the cylinder is given by $$F_s(x,t)=\begin{cases}fH(x,1-3t)&0\leq t\leq \frac{1-s}{3}\\ fH(x,3t+2s-1)&\frac{1-s}{3}\leq t\leq \frac{2(1-s)}{3}\\(x,\frac{3}{1+2s}t+1-\frac{3}{1+2s}) &\frac{2(1-s)}{3}\leq t\leq 1.\end{cases}$$

I'll leave you to sort out the details of the other homotopy $$\varphi\theta\simeq id_{M_f}$$. The point is that once you describe the composite $$\varphi\theta$$ in the form of a diagram, you immediately see what your homotopies should be. If you sketch it, you'll find that I've already written down all the ingredients you will need!

As for your additional request in the comment to Paul Frost's answer, $$X$$ includes into $$M_{fh}$$ at the $$X\times 1$$ end of the cylinder as the first part of a factorisation of $$fh$$ as a cofibration followed by a homotopy equivalence. You can see immediately that $$X\times 1$$ is fixed under the homotopy $$F$$ for all values of $$s\in I$$. Hence once you sort out the opposite direction you will see that indeed there is a homotopy equivalence of pairs $$(M_{fh},X)\xrightarrow\simeq (M_f,X)$$.

• Thanks for this very detailed answer. I did not know about this diagram-way of writing maps between mapping cylinders. Even though I feel like I understand how they are used, do you maybe have a reference for me to make sure? However, I think something went wrong with the second homotopy as the end of it maps $(x,0)$, which is identified with $f(h(x))$ to $f(x)$ which in general is not $f(h(x))$. Commented Sep 1, 2019 at 5:37
• @FriederJäckel, sorry about that. It should be fixed now. I've changed the map $\varphi$ that I originally defined, to a map homotopic to it. This makes it a lot easier to see what is going on at the end. I've rewritten from the part "To get a map in the other direction..." onwards, so let me know if you still have any queries about what now appears. Commented Sep 2, 2019 at 16:49

You have two maps $$f_k : X_k \to Y$$ and a map $$\phi : M_{f_1} \to M_{f_2}$$ such that $$\phi(y) = y$$ for $$y \in Y$$. It is well-known that $$Y$$ is a strong deformation retract of both $$M_{f_1}, M_{f_2}$$, thus the inclusions $$i_k : Y \to M_{f_k}$$ are homotopy equivalences. But $$\phi \circ i_1 = i_2$$ which proves that $$\phi$$ is a homotopy equivalence.

• I forget to mention it in the question (sorry, my mistake) but I am actually looking for a proof that the given map is a homotopy equivalence of the pairs $(M_{f\circ h},X)$ and $(M_f,X)$. Commented Sep 1, 2019 at 5:42
• Then you should edit your question (even though it has been answered by Tyrone) Commented Sep 1, 2019 at 8:14