Mapping into Hurewicz cofibrations. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:24:48Z http://mathoverflow.net/feeds/question/49289 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49289/mapping-into-hurewicz-cofibrations Mapping into Hurewicz cofibrations. Jeff Strom 2010-12-13T18:11:26Z 2010-12-13T22:34:32Z <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>