20150107.tex

\input{include.tex}

\markright{Piotr Suwara\hfill Homological algebra II: January 7, 2014\hfill}

\begin{document}
    \begin{remark}
        $C_\ast(\C)$ does not have enough projective objects.
    \end{remark}
    
    \begin{theorem}
        The sequence of functors $\{ H_i\}_{i=0}^\infty$
        gives us a~{\em universal $\delta$-functor}
        (takes short exact sequences to long exact sequences),
        i.e. if we have another sequence $T_i$ such that
        ${T_0 = H_0}$, then $\forall_i H_i^\ast= T_i^\ast$.
    \end{theorem}
    
    \begin{lemma}
        For a~given $C_\ast \in C_\ast(\C)$ there exists 
        $P_\ast \twoheadrightarrow C_\ast$
        such that $H_i(P_\ast) = 0$ for $i>0$.
    \end{lemma}
    
    \begin{remark}
        If $p+q=n$,
        let $f_{pq}:X_{nn}\to X_{pq}$ be
        defined as $d_{p+1}^h \circ \ldots \circ d_n^h
        \circ d_0^v \circ \ldots \circ d_0^v$,
        and then the Alexander-Whitney map
        $\sum_{p+q = n}f_{pq}:X_{nn} \to \bigoplus_{p+q=n}X_{pq}$
        gives a~chain homotopy equivalence 
        of $k(X_{pp})$ and $\mr{tot}(kX_{pq})$.
    \end{remark}
    
    \begin{remark}
        We may take a~projective simplicial resolution $P_\ast$ of $A$
        of degree $n>i$,
        then $L_i^s T(A) = H_{n+i}(T(P_\ast))$.
    \end{remark}
    
    \begin{theorem}
        $\deg L_i T(\bullet, n) \leq \lfloor \frac i n \rfloor$.
    \end{theorem}
    
    \begin{remark}
        Or theorem? Or proof?
        It is written that 
        \\ $T((A,n)\oplus(B,n)) = T(A,n) \oplus T(B,n) \oplus V$
        where $V$ is trivial below $2n$.
    \end{remark}
    
    \begin{proposition}
        $\forall_i \, L_i^sT$ is an additive functor.
    \end{proposition}
    
    \begin{proposition}
        Let $0 \to A \to B \to C \to 0$ be exact in $\C$.
        Then we have a~long exact sequence
        $\ldots \to L_{q+1}^s T(C) \to L_q^sT(A) \to L_q^sT(B) \to L_q^sT(C) \to \ldots$.
    \end{proposition}
    
    \begin{proposition}
        If $0 \to T' \to T \to T'' \to 0$ is an exact sequence of functors,
        then we have a~long exact sequence of functors
        $\ldots \to L_{i+1}^s T'' \to L_i^s T' \to L_i^s T \to L_i^s T'' \to \ldots$.
    \end{proposition}
    
    \begin{proposition}
        Let $U$ be an additive functor, then for any functor $T$ we have
        \\ $\Hom_{sth}(T,U) \simeq \Hom_{sth2}(L^s_0 T,U)$.
    \end{proposition}

\end{document}