Mapping into Hurewicz cofibrations. - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T07:24:48Zhttp://mathoverflow.net/feeds/question/49289http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/49289/mapping-into-hurewicz-cofibrationsMapping into Hurewicz cofibrations.Jeff Strom2010-12-13T18:11:26Z2010-12-13T22:34:32Z
<p>In Strøm's paper "The Homotopy Category is a Homotopy Category" he proves
(Lemma 4) that if $Y$ is compact and if $i:A\to X$ is a cofibration, then the induced map
$$
i_*: A^Y \to X^Y
$$
is also a cofibration. </p>
<p>The proof goes like this: he's already shown that $i:A\to X$ is a cofibration if and only
if there is a function $u: X\to I$ such that $A = u^{-1}(0)$ and a deformation
$H:X\times I \to X$ of $X$ that is constant on $A$ and pushes $U = u^{-1}([0,1))$
into $A$. Then he defines
$$
v(\alpha) = \sup \{ i\circ \alpha(Y) \}
\qquad \mathrm{and}\qquad K (\alpha, t) (y) = H(\alpha(y), t).
$$
This shows that $i_*$ is a cofibration since $v^{-1}(0) = A^Y$ and $K$ deforms
$V = v^{-1}([0,1))$ into $A^Y$.
When I try to prove that $v$ is continuous, it's very helpful to have the compactness of $Y$. </p>
<p>My question is: can the compactness hypothesis be dropped,
perhaps if we work with compactly generated spaces? </p>
<p>EDIT: Ok, I see the problem with noncompact domains.<br>
The space $I^{\mathbb{R}}$ is a good example. Functions in the set $\mathcal{U}(C,V)$
($C\subseteq \mathbb{R}$ compact, $V\subseteq I$ open) are only limited on $C$, and so can take
very large values elsewhere. You can cut this down a bit by taking intersections, and this does the job for compact domains, but you can only take finitely many. These spaces
being as nice as can be, this is a dead question.</p>