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In Allen Hatcher's book Spectral Sequence page 29 Example 1.18, enter image description here

What means "turning the map into a fibration" and convert a map into a fibration"? Given a map $f:X\to Y$, $f$ is not necessarily a fibration. How can we turn it into a fibration?

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  • $\begingroup$ For a generalization of this idea, see "model categories". $\endgroup$ – Justin Young Nov 20 '14 at 16:50
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Given a map $f:A\to B$, let $B^I$ denote the set of continuous functions $I \to B$ endowed with the compact open topology. Define $E_f=\{(a,\gamma)\in A \times B^I | \gamma(0)=f(a)\}$

One can show that the map $p:E_f \to B$, $(a,\gamma)\mapsto \gamma(1)$ is a fibration. We can view $A$ as a subspace of $E_f$ consisting of tuples $(a,\gamma)$ with $\gamma$ the constant path at $f(a)$, and we can also check that $E_f$ deformation retracts onto this subspace.

Therefore the map $f:A\to B$ factors as a composition $A\hookrightarrow E_f\twoheadrightarrow B$ of a homotopy equivalence and a fibration.

For details, refer to the subsection "Pathspace constructions" in Hatcher's Algebraic topology at page 407.

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