Any two maps to a cone space are homotopic.

Question

I have to prove that any two continuous functions to a cone space are homotopic.

Definition of cone space: If $Y$ is any topological space and $I=[0,1]$ is the closed unit interval in $\mathbb R$, then the cone space $CY$ on $Y$ is the identification space of $Y\times I$ where $\{(y,1)\mid y\in Y\}$ constitute one class and all other classes are singletons.

I have $f,g:X\to CY$ and I need to show $f$ is homotopic to $g$. I know homotopy is transitive, so I just need to show all maps are homotopic to the constant map $x\mapsto v$ where $v$ is the special point in $CY$.

I can define the homotopy, but I'm having trouble finding an elegant way to show it is continuous. Here's what I have so far:

Let $h:X\to CY$ be the constant map $h(x)=v$. Let $x\in X$. Suppose $f(x)\not=v$. Then we can write $f(x)=\overline{(f_1(x),f_2(x))}$ where $(f_1(x),f_2(x))\in Y\times I$ and $\overline{(f_1(x),f_2(x))}$ is the class of $(f_1(x),f_2(x))$ in $CY$. We can do this because as long as $f(x)\not=v$, then $f(x)$ is a class consisting of a single point. Define $F:X\times I\to CY$ by $F(x,t)=v$ if $f(x)=v$ and $F(x,t)=\overline{(f_1(x),f_2(x)(1-t) + t)}$ if $f(x)\not=v$. Then $F(x,0)=f(x)$ and $F(x,1)=v$.

Does anybody know an elegant way to show $F$ is continuous? Or otherwise, does anybody know a better way to approach this problem? Thank you!

Answer 1

Hoping to be helpful for someone...

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The key idea for this kind of problems is always the universal property of the quotient topology:

If $X$ and $Y$ are topological spaces, and $\pi : X \to Y$ is a quotient map, then for any function $f : Y \to Z$ (where $Z$ is any topological space) we have that $f$ is continuous if $f \circ \pi$ is continuous.

In particular, we have the following criterion to define continuous maps out of a quotient space:

If $X$ and $Y$ are topological spaces, and $\pi : X \to Y$ is a quotient map, then for any continuous map $g : X \to Z$ (where $Z$ is any topological space) such that $$(\forall x,x' \in X) \quad \pi(x) = \pi(x') \implies g(x) = g(x')$$ there exists a unique continuous map $\tilde g : Y \to Z$ such that $\tilde g \circ \pi = g$.

As an application of this, if $X$ is a topological space, and $A$ is a subspace of $X$, then for any continuous map $g : X \to Z$ that is constant on $A$ there exists a continuous map $\tilde g : X/A \to Z$ such that $\tilde g \circ \pi = g$, where $\pi : X \to X/A$ is the canonical projection.

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Now, I will show that if $f : X \to CY$ is a continuous map, then there exists a continuous map $F : X \times I \to CY$ such that $F(x,0) = f(x)$ and $F(x,1) = Y \times \{1\}$ for all $x \in X$. First, let $$\pi : Y \times I \to CY := (Y \times I)/(Y \times \{1\})$$ be the canonical projection, and consider the composition $$ H : (Y \times I) \times I \longrightarrow Y \times I \stackrel \pi \longrightarrow CY $$ where the first map sends $((y,t),s)$ to $(y,1-(1-t)(1-s))$ (which is clearly continuous).

Then, if we fix $s \in I$, the continuous map $g_s := H(-,s) : Y \times I \to CY$ is constant on $Y \times \{1\}$ (because $\pi(y,1) = Y \times \{1\}$ for any $y \in Y$), and so there exists a continuous map $\tilde g_s : CY \to CY$ such that $\tilde g_s \circ \pi = g_s$ (by the above observation).

Now, define $G : CY \times I \to CY$ by $(z,s) \mapsto \tilde g_s(z)$, and is easy to see that $H$ can be decomposed as $$ (Y \times I) \times I \stackrel{\pi \times \operatorname{id}} \longrightarrow CY \times I \stackrel G \longrightarrow CY. $$ Since $\pi \times \operatorname{id}$ is$^1$ a quotient map, the universal property of the quotient topology says that $G$ is continuous whenever $G \circ (\pi \times \operatorname{id}) = H$ is, which (fortunately) is our case. It follows that the composition $$ F : X \times I \stackrel{f \times \operatorname{id}} \longrightarrow CY \times I \stackrel G \longrightarrow CY $$ is our desired continuous map.

Answer 2

Hint:

A "cone space" is quite trivially contractible: you can use a straight line homotopy down to the base point.

Next any map into a contractible space is null homotopic (straight forward exercise).

And by transitivity of homotopy (homotopy is an equivalence relation), any two null homotopic maps are homotopic.

Answer 3

Here is an answer I came up with. I am basically using an explicit deformation retraction of the cone to its apex point. To do this I'll first deformation retract $Y\times I$ to $Y\times\{1\}$ and then consider the image of this under the natural projection. I believe this is precisely what you suggest to do without making use of the deformation retraction explicitly.

By the way, for future reference, this is exercise 5.6 on page 91 of Armstrong's Basic Topology.

Disclaimer 1: I know that deformation retractions are introduced later in the book, and I tried to avoid using them at first, but maybe Armstrong wants the reader to find out about them. I did come up with two other ideas that do not use deformation retractions, but I could not make too much use of them. If either of them turn out to be useful I'll add it later.

Disclaimer 2: I am not very well-versed in category theory, so pardon my diagram if it contains redundant information.

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First let me introduce some notation. Let $\alpha:=Y\times\{1\}\in CY$ be the apex point of the cone $CY$, and $c_\alpha: X\to CY, x\mapsto\alpha$ be the constant map at $\alpha$. Let $\pi: Y\times I\to CY$ be the natural projection, i.e., $\pi(y,s):=\begin{cases} \alpha&,\mbox{if } s=1\\ \{(y,s)\}&,\mbox{if } s<1\end{cases}$.

Consider the following diagram:

$$\begin{array} &Y& \cong &Y\times\{0\}& \hookrightarrow &Y\times I& \cong &(Y\times I)\times \{0\}& \hookrightarrow &(Y\times I)\times I& \stackrel{F_0}{\longrightarrow} &Y\times I& \\ &&& &\downarrow{\pi} && \downarrow{(\pi,id)} && \downarrow{(\pi,id)} &\curvearrowright& \downarrow{\pi}\\ & && &CY& \cong &CY\times\{0\}& \hookrightarrow &CY\times I& \stackrel{F}{\longrightarrow} &CY,& \end{array}$$

where

$$F_0:(Y\times I)\times I\to Y\times I,((y,s),t)\mapsto (y,(1-t)s+t)$$

and

$$F:CY\times I\to CY, (\pi(y,s),t)\mapsto \pi(F_0((y,s),t))=\pi(y,(1-t)s+t).$$

Observe that $F$ is defined so that $F\circ(\pi,id)=\pi\circ F_0$. Thus the continuity of $F_0$ will guarantee the continuity of $F$.

$F_0$ is a deformation retraction of $Y\times I$ onto $Y\times \{1\}$. Indeed, $F_0$ is clearly continuous and

\begin{align} &F_0((y,s),0)=(y,s)=id_{Y\times I}(y,s)\\ &F_0((y,s),1)=(y,1)\in Y\times\{1\}\\ &F_0((y,1),t)=(y,1)=id_{Y\times I}(y,1)\\ \end{align}

Next let us see that $F$ is well-defined. If $y,y'\in Y$, we have

$$F(\pi(y,1),t)=\pi(y,1)=\alpha=\pi(y',1)=F(\pi(y',1),t).$$

Then $F$ deformation retracts $CY=\pi(Y\times I)$ onto $\{\alpha\}=\pi(Y\times\{1\})$ by construction.

Let $f:X\to CY$ be a continuous function. Set

$$f(x):=\begin{cases}\alpha=\pi(y',1)&,\mbox{if } x\in f^{-1}(\alpha)\\ \{y_x,s_x\}=\pi(y_x,s_x)&, \mbox{if } x\not\in f^{-1}(\alpha)\end{cases}.$$

Here $y'\in Y$ is arbitrary but $y_x$ and $s_x$ (possibly) depend on $x\in X-f^{-1}(\alpha)$. Since both $F$ and $f$ are continuous, so is

$$G:X\times I\to CY, (x,t)\mapsto F(f(x),t)=F\circ(f,id)(x,t).$$

Moreover we have

\begin{align} &G(x,0)=F(f(x),0)=\begin{cases} \alpha&,\mbox{if } x\in f^{-1}(\alpha)\\ \pi(y_x,s_x)&,\mbox{if } x\not\in f^{-1}(\alpha)\end{cases}=f(x)\\ &G(x,1)=F(f(x),1)=\alpha=c_\alpha(x)\\ &\forall x\in f^{-1}(\alpha): G(x,t)=F(\alpha,t)=\alpha=f(x)=c_\alpha(x) \end{align}

Hence $f\underset{G}{\simeq} c_\alpha$ rel $f^{-1}(\alpha)$. Since homotopy relative to a subset of the domain is an equivalence relation, any two continuous function $X\to CY$ are homotopic to each other.