4
$\begingroup$

Given two maps $f:A \to B$ and $g:A \to C$, we can have the homotopy push out square

\begin{array}{rcl}A& \stackrel{f}{\rightarrow} &B\\ {\tiny g}\downarrow&& {\tiny b}\downarrow\\ C&\stackrel{a}{\rightarrow} &P\end{array}

where $a:C \to P$ and $b:B \to P$ are obvious maps.

But, I have a question for a part in some paper which I'm studying is that

Given : $ C \stackrel{gf}{\longleftarrow} A \stackrel{hf}{\longrightarrow}D$ where f is a map which has no special condition and $h:B \to D$ is the inclusion in the mapping cone sequence $ X \stackrel{k}{\longrightarrow} B \stackrel{h}{\longrightarrow}D=C_k$. So $h$ is the inclusion to the mapping cone of $k$, the mapping cone of $k$.

Obtain : We have the homotopy push out square

\begin{array}{rcl}A& \stackrel{f}{\rightarrow} & B &\stackrel{h}{\rightarrow}& D\\ {\tiny id}\downarrow&& {\tiny g}\downarrow && {\tiny b}\downarrow \\ A&\stackrel{gf}{\rightarrow} & C & \stackrel{a} {\rightarrow} & P\end{array}

where $a$ and $b$ are obvious maps.

I think that this diagram is induced by the homotopy push out of $ C \stackrel{gf}{\longleftarrow} A \stackrel{hf}{\longrightarrow}D$. It can be obtained by

\begin{array}{rcl}A& \stackrel{hf}{\rightarrow} & D\\ {\tiny gf}\downarrow&& {\tiny b}\downarrow\\ C&\stackrel{a}{\rightarrow} &P\end{array}

But, in this paper, the author said that "$P$ is the mapping cone of $gf$ so we have a mapping cone sequence $ A \stackrel{gf}{\rightarrow} C \stackrel{a}{\rightarrow} P$" where $a$ is the inclusion and $P=C_{gf}$.

Why it is ture? It is ture when $hf$ is contractible. The map $hf$ is contractible? I don't know why $P=C_{gf}$.

$\endgroup$
  • $\begingroup$ Can you give a link to the paper? It could be you're missing a hypothesis... $\endgroup$ – Dylan Wilson Apr 2 '13 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.