Rotman Algebraic Topology Exercise 1.14

Question

The question is

If CX is the cone of X, prove that $X \mapsto CX$ defines a functor $\text{Top} \to \text{Top}$ (the reader must define the behavior on morphisms) (Hint: Use Exercise 1.11)

I did exercise 11, and it says that if $f: X \to Y$ is continuous and respects an equivalence relation, i.e. if $X/\sim$, and $Y / \square$ are quotient spaces, $x \sim x'$, then $f(x) \square f(x')$. Then $f$ induces a continuous map $\bar{f}: X / \sim \to Y / \square$.

I see that if $f: X \to Y$, is continuous, then the functor $Tf: TX = CX \to TY = CY$ must be continuous. But I don't see how to do this since we don't know if the original function $f$ respects the equivalence relations.

Answer 1

It seems like the confusion comes from what $X$ and $Y$ are in exercise 11 vs the present exercise. Given a continuous map $f:X\to Y$, you have an induced map $f':X\times I\to Y\times I$ defined by $f'(x,t)=(f(x),t)$, where $I=[0,1]$. A cone $CX$ is a quotient of $X\times I$ by the relation $(x,1)\sim (x',1)$, for all $x,x'\in X$. Similarly, $CY$ is $Y\times I$ by $(y,1)\mathbin{\square} (y',1)$, for all $y,y'\in Y$.

Notice that if $(x,t)\sim (x,t')$, then either $(x,t)=(x',t')$ or $t=t'=1$. The second case is the more interesting one. Since $f'(x,1)=(f(x),1)$ and $f'(x',1)=(f(x'),1)$, we have that $f'(x,1)\mathbin{\square} f'(x',1)$. Paired with a similar statement for the first case, by exercise 11 there is an induced map $\overline{f'}:CX\to CY$. Define $Tf$ to be this $\overline{f'}$.

Answer 2

Name the canonical projections by $\pi_1: X \rightarrow X/\sim$ and $\pi_2: Y \rightarrow Y/\square$.

To prove that $\bar{f}$ is continuous, by definition we need to check the preimage of any open sets in $Y/\square$ being open in $X/\sim$.

By the definition of quotient topology, we know all open sets $V$ in $Y/\square$ are exactly those sets whose preimages in $Y$ are open.

However, $\bar{f}^{-1}(V) = \pi_1(f^{-1}\circ \pi_2^{-1}(V))$, so by definition of the quotient topology of $X/\sim$, $\bar{f}^{-1}(V)$ is open. This proves your statement.

EDIT (for the comment. I am not able to comment before my reputation grows more than 50): from any $f^0: X\rightarrow Y$ you have a naturaly $:XxI \rightarrow YxI$. Then apply my argument above to this $f$.