# (Co-)fibrations in Top and CGWH

Suppose that you have a map $$i: A\rightarrow X$$ between CGWH (compactly generated weakly Hausdorff) spaces. It is true that if it is a cofibration in CGWH (the category of CGWH spaces), then it is a cofibration in Top (the category of topological spaces)?

• Interesting question. You certainly know Neil Stricklands's paper "The category of CGWH spaces" neil-strickland.staff.shef.ac.uk/courses/homotopy/cgwh.pdf. It is easy to see that the product of weakly Hausdorff spaces it weakly Hausdorff. Hence by Strickland 2.6, if $X$ is CGWH, then so is $X \times I$. Using Strickland 1.10 we see that if a map $i : A \to X$ in CGWH is a cofibration in CG, then it is also one in Top. However, there is a still a gap: If $i$ is a cofibration in CGWH, is it one in CG? – Paul Frost Dec 6 '18 at 16:20
• If $i(A)$ is closed in $X$, then $i$ is a cofibration in Top iff $Z = X \times 0 \cup i(A) \times I$ is a retract of $X \times I$. It suffices to know that $i$ is a cofibration in CGWH to show that $Z$ is a retract of $X \times I$: $Z$ is closed in $X \times I$, hence a CGWH space. – Paul Frost Dec 6 '18 at 19:54
• @PaulFrost Actually, $i$ is a cofibration in Top iff the canonical map $l: M_i\to X\times I$ has a retract iff $Z$ is a retract of $X\times I$, so the closedness assumption is not necessary. (However, it can be shown that a cofibration in CGWH is always a closed embedding.) But knowing that a map is cofibration in CGWH is indeed enough to conlude that $l$ has a retract (because the pushout of a diagram in which at least one of the maps is a closed embedding is in CGWH). Thank you! – Gregg Dec 8 '18 at 19:40
A map $$i: A\to X$$ is a cofibration in Top iff the canonical map $$l: M_i\to X\times I$$ has a retract iff $$Z=X\times \{0\}\cup i(A)\times I$$ is a retract of $$X\times I$$. Knowing that a map is cofibration in CGWH is enough to conlude that $$l$$ has a retract, because the pushout of a diagram in which at least one of the maps is a closed embedding is in CGWH.