Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $Z$ be a $\Delta$-generated space (a colimit of simplices -not sure that this hypothesis is important but it is the framework I am working in). The set of continuous maps $Z^{[0,1]}$ from $[0,1]$ to $Z$ is equipped with the $\Delta$-kelleyfication of the compact-open topology (the internal hom of the category, see below).

QUESTION : Suppose that the evaluation map at $0$ from $Z^{[0,1]}$ to $Z$ satisfies the right lifting property with respect to any monomorphism of $\Delta$-generated spaces (i.e. injections). Then I suspect that $Z$ must be discrete. Any counter-example ?

Concerning $\Delta$-generated spaces, a short bibliography, just in case that it is important : 1. Notes on Delta-generated spaces by Dugger : http://math.uoregon.edu/~ddugger/delta.html 2. The proof that they assemble to a locally presentable category : A convenient category for directed homotopy by Fajstrup-Rosicky : http://www.tac.mta.ca/tac/volumes/21/1/21-01abs.html 3. A survey of their properties : Section 2 of Homotopical interpretation of globular complex by multipointed d-space http://www.tac.mta.ca/tac/volumes/22/22/22-22abs.html

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

$\newcommand{\into}{\hookrightarrow}$It seems that if $Z$ has the indiscrete topology, then the evaluation map $ev_0 : Z^I \to Z$ has the right lifting property with respect to any map. That provides a simple counter-example to the question.

So I will assume for the remainder of my answer that $Z$ is $T_1$, i.e. singleton subsets are closed in $Z$. I will show that $Z$ is necessarily discrete if $ev_0 : Z^I \to Z$ has the right lifting property with respect to the inclusion $J \into I$, where $J={}]0,1]$ is a non-closed interval.

By the way, I do not know how to deal with more general classes of spaces, as the argument below uses very strongly the fact that the space $Z$ is generated by simplices, which are path connected spaces.

Claim 0: If $f,g:J\to Z$ are homotopic maps, and $g$ extends to $I$, then $f$ also extends to $I$.

This follows immediately by applying the assumed right lifting property of $ev_0 : Z^I \to Z$ to any homotopy from $g$ to $f$, seen as a map $J\to Z^I$.

Claim 1: Any map $J\to Z$ extends to a map $I\to Z$.

Any map $J\to Z$ is homotopic to a constant map, since $J$ is contractible. The statement now follows from claim 0.

Claim 2: Any continuous map from a simplex to $Z$ is constant.

Since simplices are path connected, it suffices to show that any map $f:I\to Z$ is constant. Let $x,y\in I$.

Consider the continuous map $h:J\to I$ given by $$ h(t) = \frac 1 2 + \frac 1 2 \sin\Bigl(\frac 1 t\Bigr) $$ For any neighbourhood $U$ of zero in $I$, there exist points $a,b\in J\cap U$ such that $h(a)=x$ and $h(b)=y$.

By claim 1, the map $f\circ h$ extends to a continuous map $g:I\to Z$. By our choice of $h$, within any neighbourhood of zero in $I$ there exist points $a$, $b$ such that $g(a)=f(x)$, and $g(b)=f(y)$. Therefore, $g(0)\in \overline{\{f(x)\}}$ and $g(0)\in \overline{\{f(y)\}}$. Since $Z$ is $T_1$, it follows that $f(x)=g(0)=f(y)$.

Conclusion: $Z$ is a discrete space.

Indeed, since $Z$ is $\Delta$-generated, a subset $S$ of $Z$ is open if the inverse image $f^{-1}(S)$ is open for any map $f$ from a simplex to $Z$. Claim 3 then implies that singleton subsets of $Z$ are open, i.e. $Z$ is discrete.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.