# Fiber sequences in proper model categories

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) $$\begin{array}{ccccc} X&\to &Y&\to&Z \newline \downarrow&&\downarrow&&\downarrow\newline F&\to&E&\to&B \end{array}$$ in $Ho \mathcal{C}$ commutes where $F$ is the point set fiber, the vertical maps are all isomorphisms and the diagram (2)

$$\begin{array}{ccc} X\times\Omega Z&\to& X\newline \downarrow && \downarrow\newline F\times\Omega B&\to& F \end{array}$$ 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.

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