Fiber sequences in proper model categories - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:00:48Z http://mathoverflow.net/feeds/question/70901 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70901/fiber-sequences-in-proper-model-categories Fiber sequences in proper model categories Rolf N. 2011-07-21T13:29:40Z 2011-07-28T08:48:00Z <p>I am confused about the notion of a fiber sequence (or dually a cofiber sequence) in a general pointed and proper model category $\mathcal{C}$.</p> <p>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.</p> <p>A <em>fiber sequence</em> (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) $$\begin{array}{ccccc} X&amp;\to &amp;Y&amp;\to&amp;Z \newline \downarrow&amp;&amp;\downarrow&amp;&amp;\downarrow\newline F&amp;\to&amp;E&amp;\to&amp;B \end{array}$$ in $Ho \mathcal{C}$ commutes where $F$ is the point set fiber, the vertical maps are all isomorphisms and the diagram (2)</p> <p>$$\begin{array}{ccc} X\times\Omega Z&amp;\to&amp; X\newline \downarrow &amp;&amp; \downarrow\newline F\times\Omega B&amp;\to&amp; F \end{array}$$ commutes where the vertical morphisms are induced by (1) and the lower horizontal one comes from the previous remark. An <em>equivalence of two fiber sequences</em> $(X\to Y\to Z,\varphi)$ 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 $(X'\to Y'\to Z',\varphi')$.</p> <p>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.</p> <p>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.</p>