Which notion of equivalence is being used here?

Question

I'm reading Homotopy associativiy of H-spaces I by Stasheff and in the proof of Proposition 2 he says that any map is equivalent to a fibring (in modern day language I think this is either a fibration or a fiber bundle). He refers to Lemma 13 of this paper by Nomura, where he states that any map $f:X\to Y$ is equivalent to a fibring. The fibring is of the form $p:Z_f\to Y$ where $Z_f$ is constructed in the proof.

I would like to know in what sense these maps are equivalents. The map $f$ need not be a fibring so this is not an equivalence of fibrings. A homotopy equivalence wouldn't make sense either because the domains are different. I haven't found in any of the linked papers any definition of this equivalence, so does anyone knows what they're talking about?

According to Stashef there should be a map $X\to Z_f$ commuting with $f$ and $p$, so maybe they mean that $f$ is homotopy equivalent to the composition of $p$ with this map.

Answer 1

By fibring Stasheff means Hurewicz fibration. That is a map $p:E\rightarrow B$ which has the homotopy lifting property with respect to all spaces.

By equivalent to Stasheff means the following. Given any map $f:X\rightarrow Y$ there is a Hurewicz fibration $p_f:W_f\rightarrow Y$ and inverse homotopy equivalences $$X\xrightarrow{i} W_f\xrightarrow{q} X$$ such that $$p_f\circ i=f,\qquad \qquad f\circ q\simeq p_f.$$ It is in the sense of the first equation that $f$ is equivalent to $p_f$.

As a model for the space it is standard to take $$W_f=\{(x,l)\in X\times Y^I\mid f(x)=l(0)\}$$ with $$p_f(x,l)=l(1).$$ Here $Y^I$ is the space of all maps $I\rightarrow Y$ in the compact-open topology. The map $i$ sends $x$ to the pair $(x,c_{f(x)})$, where $c_{f(x)}$ is the constant path at $f(x)$. The map $q$ is the projection $(x,l)\mapsto x$.

See, eg. M. Arkowitz's book Introduction to Homotopy Theory $\S$3.3 for details.