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Let $A^\bullet$ be the complex:

$\cdots \rightarrow A^{n-2} \xrightarrow{d^{n-2}} A^{n-1} \xrightarrow{d^{n-1}} A^{n} \xrightarrow{d^{n}} A^{n+1} \xrightarrow{d^{n+1}} A^{n+2} \xrightarrow{d^{n+2}} \cdots$,

consider the truncation $\tau_{\leq n}A^{\bullet}$:

$\cdots \rightarrow A^{n-2} \xrightarrow{d^{n-2}} A^{n-1} \xrightarrow{d^{n-1}} \ker d^{n} \rightarrow 0 \rightarrow 0 \rightarrow \cdots$,

consider a resolution for $\tau_{\leq n}A^{\bullet}$: the quasi-isomorphism $f_n: P^{\bullet}_{n} \rightarrow \tau_{\leq n}A^{\bullet}$

enter image description here.

Now I'm reading a paper and I got to this part that says:

"... let $g: P^{\bullet}_{n} \rightarrow \tau_{\leq n + 1}A^{\bullet}$ be the chain map induced by $f_n$...".

But I wanted to be perfectly clear and ask, what is this chain map $g$? Is it

enter image description here

If so, what would be the map $q$?

share|improve this question
    
imgur warning. Can we have the source code of the diagrams as a backup? –  darij grinberg Jan 19 '13 at 0:45
    
Hey, can you see the diagrams? I can see them on my computer just fine. They're not on any hosting site, I uploaded them directly from my computer. –  Mario Carrasco Jan 19 '13 at 0:52
2  
$f_n^n$ lands in $\ker d^n$ which is contained in $A^n$, so the without information about the context there are few options (or none, even!) apart from lettng $q$ be the composition of $f_n^n$ with the inclusion $\ker d^n\to A^n$. –  Mariano Suárez-Alvarez Jan 19 '13 at 5:52
    
Thank you @Mariano Suarez-Alvarez –  Mario Carrasco Jan 22 '13 at 19:28

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