# Talk:Chain duality III (Ex)

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The most interesting part is to check equivariance, say for objects $M\in\mathbb A$$; The most interesting part is to check equivariance, say 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 T'_{F(M),F(M)}(F (\varphi) \circ G(M))=e'_{F(M)}\circ T'G(M) \circ T'F(\varphi) and F(T_{M,M}\varphi)\circ G(M)=F(e_M)\circ FT\varphi\circ G(M). This follows from the commutative diagram \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. M\in\mathbb A$.

Let $\varphi:TM\to M$$\varphi:TM\to M$ be an element of $M\otimes_{\mathbb A}M$$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$$G$ and the second one by definition of a functor of categories with chain duality.