Showing that every continuous map between two topological spaces $X$ and $Y$ induces a continuous map between their cones $C(X)$ and $C(Y)$

Question

I’d like to know how to show that every continuous map between two topological spaces $X$ and $Y$ induces a continuous map between their cones $C(X)$ and $C(Y)$. I was thinking it could be a good idea to try to use the universal properties for the quotient and/or the product topology but I’m not sure how to proceed. I think a continuous map between spaces induces a continuous map between quotient spaces, is that the way to go?

Answer 1

If $f: X \to Y$, then $C(X)$ is the quotient of $X \times [0,1]$ by identifying the subset $X \times \{1\}$ to a point and likewise for $C(Y)$.

It's clear that $f$ defines a map $f \times 1: X \times [0,1] \to Y \times [0,1]$ by $(f \times 1)(x,t)=(f(x),t)$ (continuous by standard results on the product topology) and this clearly has the property that the points we are going to identity for $C(X)$ all map into $Y \times \{1\}$, so that $C(f): C(X) \to C(Y)$ defined by $C(f)([(x,t)]) = [(f(x),t)]$ is well-defined and continuous by standard results on quotient maps.

Answer 2

Lemma 1. Let $f:X\to Y$ and $f':X'\to Y'$ be two continuous map. Define $F:X\times X'\to Y\times Y'$ by $F(x,x'):=\big(f(x),f'(x)\big)$. Then $F$ is continuous.

Proof. It is enough to show that preimage of every open cube $U\times V$ is open, where $U\subseteq Y$ is open and $V\subseteq Y'$ is open. You can easily verify that $F^{-1}(U\times V)=f^{-1}(U)\times f'^{-1}(V)$ which is open since both $f$ and $f'$ are continuous. $\Box$

Lemma 2. Let $f:X\to Y$ be a continuous map, $\sim$ and equivalence relationship on $X$ and $\sim'$ and equivalence relationship on $Y$ such that if $x\sim y$ then $f(x)\sim' f(y)$. Then the induced map $$F:X/\sim \to Y/\sim'$$ $$F([x]_\sim):=[f(x)]_{\sim'}$$ is well defined and continuous.

Proof. It is clearly well defined because $x\sim y$ implies $f(x)\sim' f(y)$. Now let $\pi_X:X\to X/\sim$ and $\pi_Y:Y\to Y/\sim'$ be respective quotient maps. Then $\pi_Y\circ f:X\to Y/\sim'$ is of course continuous. Since $x\sim y$ implies $\pi_Y\circ f(x)=\pi_Y\circ f(y)$ then we can apply the universal property of quotient maps with respect to $\pi_X$ to conclude that there's a unique continuous map $G:X/\sim\to Y/\sim'$ such that $G\circ\pi_X=\pi_Y\circ f$. The last equality is just another way of saying that $G([x]_\sim)=[f(x)]_{\sim'}$, i.e. that $G=F$. $\Box$

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Now $C(X)=(X\times [0,1])/(X\times \{1\})$. So first, if $f:X\to Y$ is continuous then it induces a continuous map $F:X\times[0,1]\to Y\times[0,1]$ by $F(x,t):=\big(f(x),t\big)$. This is our Lemma 1 applied to $f$ and the identity $id:[0,1]\to[0,1]$, $id(t)=t$.

Now given this $F$ map all we need to see is that it maps equivalent elements to equivalent elements to apply Lemma 2. But this is straightforward since we collapse only elements with second coordinate $1$, and $F$ does not change the second coordinate.

All in all, our continuous $f:X\to Y$ induces a continuous $F:C(X)\to C(Y)$ map given by $F([x,t]):=[f(x),t]$.

Answer 3

Note that $CX= \frac{X\times I}{X\times \{0\}}$. Suppose $f:X\rightarrow Y$ is continuous. Then $f$ induces a continuous map $f\times id: X\times I \rightarrow Y\times I$. $Cf$ is simply the quotient map of $f\times id$. That is, $(Cf)([x,t])=[f(x),t]$. We need to establish a few things.

1. $Cf$ is well defined:

Either $(x,t)=(x',t')$ or $(x,t),(x',t')\in X\times \{0\}$ . In the first case, obviously, $[f(x),t]=[f(x'),t']$ . If on the other hand $(x,t),(x',t')\in X\times \{0\}$ , then, $t=t'=0$ and so $(f(x),0),(f(x'),0)\in Y\times \{0\}$ . Hence $[f(x),t]=[f(x'),t']$

2. $Cf$ is continuous:

Note that $Cf\circ q_{X\times I}=q_{Y\times I}\circ (f\times Id)$, where $q_{X\times I}$ is the quotient map $X\times I\rightarrow CX$ and $q_{Y\times I}$ is defined similarly.

A remark on notation: $[a,b]$ really is $[(a,b)]$. But for the sake of clarity, I have chosen to omit the brackets.