I am confused about the notion of a fiber sequence (or dually a cofiber sequence) in a general pointed and proper model category $\mathcal{C}$.

Following Hovey, we can define, like in topology, a map $\phi: F\times\Omega B\to F$ for a fibration $p:E\to B$ of fibrant objects in $\mathcal{C}$ where $F=p^{-1}(*)$ is the point set fiber.

A fiber sequence (after Hovey) is a diagram $X\to Y\to Z$ in the homotopy category $Ho \mathcal{C}$ together with a morphism $\varphi: X\times\Omega Z\to X$ in $Ho \mathcal{C}$ with the following property: There exists a fibration $p:E\to B$ of fibrant objects such that the diagram (1) \begin{equation} \begin{array}{ccccc} X&\to &Y&\to&Z \newline \downarrow&&\downarrow&&\downarrow\newline F&\to&E&\to&B \end{array} \end{equation} in $Ho \mathcal{C}$ commutes where $F$ is the point set fiber, the vertical maps are all isomorphisms and the diagram (2)

\begin{equation} \begin{array}{ccc} X\times\Omega Z&\to& X\newline \downarrow && \downarrow\newline F\times\Omega B&\to& F \end{array} \end{equation} commutes where the vertical morphisms are induced by (1) and the lower horizontal one comes from the previous remark. An equivalence of two fiber sequences $(X\to Y\to Z,\varphi)$ and $(X'\to Y'\to Z',\varphi')$ is defined with the same diagrams (1) and (2) by replacing $(F\to E\to B,\phi)$ by $(X'\to Y'\to Z',\varphi')$.

One can show that for an arbitrary morphism $f$ in $\mathcal{C}$, the diagram $hofib(f)\to Y\xrightarrow{f} Z$ can be made into a fiber sequence by specifying a certain map $\varphi$. My question is sloppy phrased as: Does one have a choice for the operation $\varphi$? Please note that I require $\mathcal{C}$ to be proper.

Let me make this more precise. If $(X\to Y\xrightarrow{f} Z,\varphi)$ is a fiber sequence, then $X$ is weakly equivalent to $hofib(f)$ by diagram $(1)$. If $Z$ is fibrant, the rightmost vertical arrow in (1) can be chosen as the identity which means in particular that condition (2) becomes obsolete, i.e. all fiber sequences $(X\to Y\to Z,?)$ are equivalent, if this observation is correct. Is this also true if $Z$ is not fibrant, i.e. is there (up to equivalence) a unique fiber sequence comming from $X\to Y\to Z$? Thank you.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.