Commutativity up to homotopy implies strict commutativity, for lifting problems

Suppose we have a commutative diagram $$\require{AMScd}$$ $$\begin{CD} A @>>> X \\ @VVV & @VVV \\ W @>>> Y\\ \end{CD}$$ where the map $$A\rightarrow W$$ is a cofibration and the map $$X\rightarrow Y$$ is a fibration. Suppose also that there exists a map $$W\rightarrow X$$ that makes the diagram commute up to homotopy. Is then true that we can find a map $$W\rightarrow X$$ that makes the diagram strictly commute?

• What setting are you working in — an arbitrary Quillen model category? a specific notion of “fibration”/“cofibration” in $Top$, or some well-behaved subcategory thereof? – Peter LeFanu Lumsdaine Oct 9 '18 at 11:49
• I'm working in Top category, more precisely $(W, A)$ is a CW-pair, but I'm not sure whether this hypothesis is necessary. – Diego95 Oct 9 '18 at 12:11
• I believe this follows (in a general model category) from Proposition A.2.3.1 in Higher Topos Theory - the statement only covers the case where $Y$ is the final object, but the more general case follows via working in the overcategory over $Y$. – dhy Oct 9 '18 at 14:47

I believe the answer is yes. The kind of lift you're asking about was studied extensively in the paper "On Fibrant objects in model categories" by Valery Isaev. Apply Proposition 3.4, with $$I = \{A \to W\}$$. Then, because $$f:X\to Y$$ is a fibration, it has RLP with respect to $$J_I$$, because $$J_I$$ consists of trivial cofibrations. Hence, Proposition 3.4 says that, having RLP up to relative homotopy with respect to $$I$$ implies $$f$$ has RLP with respect to $$I$$, as you desire.
• Let call $g:W\rightarrow X$ the map that makes the square commute up to homotopy (it's not difficult to show that I can find such a $g$ that makes the top triangle strictly commute). Now it seems to me that in order to apply the proposition 3.4 I need that the homotopy of the bottom triangle must be relative to A. How can I suppose this? – Diego95 Oct 9 '18 at 16:45