A sequence which is homtopy equivalent to a fibration is a fibration

Question

Suppose we have a fibration $F\to E \to B$ and a sequence of maps $F'\to E' \to B'$ and suppose we have a map between these two sequences, which commutes up to homotopy and which is pointwise a homotopy equivalence (i.e each map $\bullet\to\bullet', \bullet\in\{F,E,B\}$ in the homotopy commutative map is a homotopy equivalence).

Is there any chance that then $F'\to E'\to B'$ is a homotopy fibration? We will obviously have the homotopy long exact sequence as if it were a homotopy fibration, but will it have the homotopy lifting property?

thanks in advance for your help and advice.

Answer 1

For an explicit counter-example, you can consider something like $$\require{AMScd}\begin{CD} \{1\} @>{\subseteq}>> I @>{=}>> I\\ @V{=}VV @V{i}VV @V{=}VV\\ \{\ast\} @>{\subseteq}>> CX @>{p}>> I \end{CD}. $$ Here, $CX=X\times I/X\times\{1\}$ is the cone over some space $X$, the map $i$ is given by $t\mapsto [x_0,t]$ for some basepoint $x_0\in X$ and the map $p$ is given by $[x,t]\mapsto t$. The squares strictly commute, the vertical maps are homotopy equivalences (the outer vertical maps are even identities) and the top row is clearly a fibration sequence, yet the bottom row is not necessarily so. Indeed, a fibration over a path-connected base has homotopy-equivalent fibers, yet the fiber of $p$ over $1$ is the cone point $\{\ast\}$ whereas the fiber of $p$ over any $t<1$ is a homeomorphic copy of $X$, so it suffices to choose a non-contractible $X$ to get a counter-example.

Answer 2

Let $B$ be a CW-complex, and consider the model structure on the slice category $\mathsf{Top}/B$ induced from one on $\mathsf{Top}$. Suppose given any map $p'\colon E'\to B$, with $E'$ a CW-complex as well. We can fibrantly replace $p'$ in $\mathsf{Top}/B$, giving us a fibration $p\colon E\to B$ and a trivial cofibration $i\colon E'\to E$ such that $p'=pi$. Since $E$ is then also a CW-complex and $i$ is a weak equivalence, we find that $i$ is a homotopy equivalence in $\mathsf{Top}/B$, i.e. a fiberwise homotopy equivalence. In particular, we get a homotopy equivalence $F'\to F$ between the generic fibers of $p$ and $p'$. This homotopy equivalence $F'\to F$ commutes with $i$ in an appropriate sense (for instance, if there is a preferred point $b\in B$, we can just take $F'=E'_b$ and $F=E_b$, then we have strict commutativity). But $p'$ need not be a fibration in general. This means that in the situation of your question, it is generally not true that $F'\to E'\to B'$ will be a fibration sequence.