TITLE: Commutativity up to homotopy implies strict commutativity, for lifting problems
QUESTION [7 upvotes]: 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?

REPLY [7 votes]: 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.
There may be older or more elementary proofs of this fact, but Isaev's paper is what sprung to mind. Also, it works in much more general settings than Top, and you might find lots of useful facts in it, for whatever you're working on.