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Is it true that if $$F\to E\to B$$ is a homotopy fibration and $F\to X$ is nullhomotopic, then there exists a map $B\to X$ making the diagram commutative?

Thank you.

Haldot
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  • Is $E\to X$ an inclusion as indicated in the picture? $E\to B$ need not be injective... – red whisker May 13 '23 at 01:28
  • Well, it is in my case of specific spaces (and in this case $E\to B$ indeed is an inclusion). Perhaps, I should have made all the arrows regular in order for the question to be correct. Thanks. – Haldot May 13 '23 at 14:05
  • Then I'm not sure what you mean by homotopy fibration. If $E\to B$ has the homotopy lifting property and $E$ is a subspace of $B$ (in which case it contains every path component it intersects; in most cases $B$ is path connected) $F$ is a singleton (preimage of the basepoint?), so $F\to X$ is always null homotopic. If you demand the map $B\to X$ for $X = E$, that's the same as asking for $E$ to be a retract of $B$ in which case we can always extend maps out of $E$ by composition with the retraction.. – red whisker May 14 '23 at 01:57
  • By homotopy fibration I basically mean that $E\to B$ is not a fibration itself, but it factors as $E\to\widetilde{E}\to B,$ where the first arrow is a homotopy equivalence, and the second arrow is a fibration. $F$ is the fiber of the second arrow. They are also called homotopy sequences. I kinda tried to describe what I know about them (not much) here. – Haldot May 14 '23 at 02:30
  • Also I know that they do have some curious lifting properties (like here, so I really hope that answer to the question is positive. It would be useful for the problem I am trying to work on. Btw, another source on them. – Haldot May 14 '23 at 02:34
  • (In my mind) fibrations have a "big" space $E$ and a "small" space $B$. When talking about homotopy fibration, "inclusion" feels like the wrong word because "there's something in between" and maps are defined up to homotopy anyway. Since the idea is to "replace" any map by a fibration, my initial thought was to prove the result for actual fibrations, in which case the homotopy fibre is homotopy equivalent to the "inverse image" fibre, and one composition being nullhomotopic makes the other too. Cofibrations are better suited to handle maps out of spaces. – red whisker May 14 '23 at 12:53

1 Answers1

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Let $$B = \{(x, \sin(1/x)):x\in(0, 1]\}\cup\{(-x, \sin(1/x)):x\in (0, 1]\}\cup \{(0, y):y\in [-1, 1]\}$$ and write $C_1$ to be the left "sine part" and $C_2$ the right and $L$ the "line part". $B$ is connected but has three path components: $C_1, C_2, L$.

Take $E = C_1\cup C_2$. Since it is a union of path components, the inclusion $E\hookrightarrow B$ is a fibration because any homotopy $Z\times I\to B$ with one end in $E$ must land in $E$. The fibre (with basepoint $(1, \sin 1)$ say) is a singleton.

There is no retraction $B\to E$ because $B$ is connected but $E$ is not. In fact, any map $f\colon B\to E$ lands in $C_1$ or $C_2$, so the composition $E\hookrightarrow B\xrightarrow{f}E$ lands in $C_1$ or $C_2$ and hence can't even be homotopic to the identity map, i.e., there's no dotted arrow making the diagram commute up to homotopy.

More generally, consider the covering map $S^1\to S^1$ given by squaring (this is a 2-1 covering with a finite fibre and the inclusion of the fibre is null homotopic because $S^1$ is path connected). There is no map making the diagram commute (even up to homotopy) for $E = X = B = S^1$, $E\to X$ identity, $E\to B$ squaring.

red whisker
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