Talk:Chain duality III (Ex)

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FT^2M\ar[r]^{Fe_M}&
FT^2M\ar[r]^{Fe_M}&
F(M)
F(M)
}.$$
+
}$$
+
as the first square commutes by naturality of $G$ and the second one by definition of a
+
functor of categories with chain duality.
</wikitex>
</wikitex>

Revision as of 11:34, 1 June 2012

We check this for objects M\in\mathbb A.

Let \varphi:TM\to M be an element of M\otimes_{\mathbb A}M.

We have to check the equality of

\displaystyle T'_{F(M),F(M)}(F (\varphi) \circ G(M))=e'_{F(M)}\circ T'G(M) \circ T'F(\varphi)
and
\displaystyle F(T_{M,M}\varphi)\circ G(M)=F(e_M)\circ FT\varphi\circ G(M).

This follows from the commutative diagram

\displaystyle \xymatrix{ T'F(M)\ar[r]^{T'F(\varphi)} \ar[d]_{G(M)} &  T'FT(M) \ar[r]^{T'G(M)} \ar[d]_{G(TM)} & T'^2F(M) \ar[d]_{e'_{F(M)}}\\ FT(M)\ar[r]^{FT\varphi} & FT^2M\ar[r]^{Fe_M}& F(M) }

as the first square commutes by naturality of G and the second one by definition of a functor of categories with chain duality.

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