1
$\begingroup$

In the cone construction (section 9.3.1) of tom Dieck's book Algebraic Topology, the author defines the following map $$q:\Delta ^{n-1}\times I\rightarrow \Delta ^n,\;\;\;((\mu_0,\dots ,\mu_{n-1}),t)\mapsto (t,(1-t)\mu_0,\dots ,(1-t)\mu_{n-1})$$ On a contractable space $X$, let $h:X\times I\rightarrow X$ be a homotopy from the identity to a constant map. The author now states

Given $\sigma :\Delta ^{n-1}\rightarrow X$, there exists a unique simplex $s(\sigma)=s\sigma :\Delta ^n\rightarrow X$ such that $h\circ (\sigma \times \operatorname{id})=s(\sigma)\circ q$, since $q$ is a quotient map.

A quotient map is defined in the book as a surjective map between topological spaces $q:X\rightarrow Y$ with the property that $U$ is open $\iff$ $q^{-1}(U)$ is open. How does the statement follow?

$\endgroup$
2
  • $\begingroup$ The correct name of the author is tom Dieck (tom is part of the surname). $\endgroup$
    – johndoe
    Feb 11, 2015 at 9:58
  • $\begingroup$ @johndoe corrected, thank you. $\endgroup$
    – user153312
    Feb 11, 2015 at 9:59

1 Answer 1

1
$\begingroup$

In general, if $q : X \rightarrow Y$ is a quotient map and $f : X \rightarrow Z$ is a continuous map such that $q(x) = q(y)$ implies $f(x) = f(y)$ then there exists a unique continuous map $g : Y \rightarrow Z$ such that $f = g \circ q$.

In your case we have $q(\mu,t) = q(\mu',t)$ if and only if $t = t'$ and $\mu = \mu'$ or $t = t' = 1$. So every map from $\Delta \times I$ which is constant on $\Delta \times \{1\}$ factors through $q$.

Now note that the map $h \circ (\sigma \times \text{id}) : \Delta^{n-1} \times I \rightarrow X$ maps all points in $\Delta^{n-1} \times \{1\}$ to a single point because $h|_{X \times \{1\}}$ is constant. Hence it factors through $q$, that is, there exists $s(\sigma)$ with the desired property.

Edit: to clarify the connection between quotient maps as defined by Dieck and equivalence relations

If $q : X \rightarrow Y$ is a surjective map then you have the equivalence relation $R_q = \{(x,y) \:|\: q(x) = q(y)\}$ on $X$. We can then construct the quotient space $X/R_q$ consisting of the equivalence classes with respect to $R_q$ and get a topology on $X/R_q$ by defining a subset of $X/R_q$ as open if its preimage under $\pi: X \rightarrow X/R_q$ is open. We then have an induced continuous map $i: X/R_q \rightarrow Y$ which is a bijection and satisfies $i \circ \pi = q$. Now, if $q$ is a quotient map then this bijection is a homeomorphism because for open $U \subseteq X/R_q$ we have that $q^{-1}(i(U)) = \pi^{-1}(U)$ is open and hence $i(U)$ is open.

$\endgroup$
0

You must log in to answer this question.