Two-point desuspension for augmented chain complexes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:43:16Z http://mathoverflow.net/feeds/question/98540 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98540/two-point-desuspension-for-augmented-chain-complexes Two-point desuspension for augmented chain complexes? Harry Gindi 2012-06-01T05:05:16Z 2012-06-01T05:10:16Z <p>Let $X$ be a chain complex augmented over $\mathbf{Z}$ with augmentation $\varepsilon_X:X_0 \to \mathbf{Z}$. We define <code>$[1](X)$</code> to be the $\mathbf{Z}$-augmented chain complex such that <code>$[1](X)_0 = \mathbf{Z}^2$</code> (on a basis $p_0,p_1$), and such that <code>$[1](X)_{k+1}=X_k$</code> for all $k$. The augmentation map <code>$\varepsilon_{[1](X)}:[1](X)_0\to \mathbf{Z}$</code> sends the basis elements $p_0,p_1$ to $1\in \mathbf{Z}$. For the boundary map <code>$\partial:[1](X)_1 \to [1](X)_0$</code>, we let $\partial(a)=\varepsilon_X(a)(p_1-p_0)$. For all other boundary maps, we let <code>$\partial:[1](X)_{k+2} \to [1](X)_{k+1}$</code> be the map $\partial:X_{k+1} \to X_k$. </p> <p>This functor <code>$[1](-)$</code>, the two-point suspension, gives an initial-object preserving functor <code>$P:\mathbf{AugCh}\to ([1](0)\downarrow \mathbf{AugCh})$</code>. </p> <p>Then my question: Is the functor <code>$[1](-)$</code> a "parametric left adjoint"? That is, is the functor $P$ a left adjoint? If this is the case, is there any explicit way to to construct the right adjoint in terms of chain complexes? </p>