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--- blog:2023-12-06 [2023/12/07 02:41] – pzhou
+++ blog:2023-12-06 [2023/12/07 15:13] (current) – [Mittag-Leffler] pzhou
@@ Line 38: / Line 38: @@
 Now, is this map $H^k (\varprojlim_n X_n) \to \varprojlim H^k(X_n)$ surjective?
 We consider taking two exact seq
-$$ 0 \to B^k*(X_\infty) \to Z^k(X_\infty) \to H^k(X_\infty) \to 0 $$
+$$ 0 \to B^k(X_\infty) \to Z^k(X_\infty) \to H^k(X_\infty) \to 0 $$
-$$ 0 \to \lim B_n^k \to \lim Z_n^k \to \lim H_n^k \to 0 $$
+$$ 0 \to \lim B^k(X_n) \to \lim Z^k(X_n) \to \lim H_n^k \to 0 $$
 the last part is because $B_n^k$ are ML in $n$, hence the proj lim SES is also right exact.
   * we have snake lemma
   * middle column isom implies right column surjective
-  *
+  * however, $B^k(X_\infty) =Im(X^{k-1}_\infty) \to \lim_n B^k(X_n)$ may not be surjective without further assumption. This is because $B^k$ is like taking cokernel of a map, and $\lim$ is taking projective lim, they don't commute. If we assume $H^{k-1}(X_n)$ is ML, then we know $Z^{k-1}(X_n)$ is ML. Then, we have
+$$ 0 \to Z^{k-1}(X_n) \to X^{k-1}_n \to B^k(X_n) \to 0 $$
+then we use the proposition that the first term is ML, then the projective limit preserve the SES, and we get
+$ B^k(X_\infty) \cong \lim B^k(X_n) $
+----
+But why do we care about these ML stuff? Why do we care about the non-characteristic deformation lemma?