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 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. 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.

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.

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 a map $\varphi: F\times\Omega B\to F$ 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.

One can show that a diagram $hofib(f)\to Y\xrightarrow{f} Z$ is (more precisely:'can be made into', by specifying an 'operation' $\varphi$) a fiber sequence for an arbitrary morphism $f$ in $\mathcal{C}$. My question is sloppy phrased as: In how far is the action $\varphi$ important? Has one a choice?

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. This should imply, that, if $Z$ is fibrant, the morphism $f:Y\to Z$ in $\mathcal{C}$ is part of an unique fiber sequence $hofib(f)\to Y\to Z$. Is this correct? What happens, when $Z$ is not fibrant?

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 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. 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.