Talk:Chain duality III (Ex)

We check this for objects $M\in\mathbb A$$<wikitex>; We check this for objects $M\in\mathbb A$. Let $\varphi:TM\to M$ be an element of $M\otimes_{\mathbb A}M$. We have to check the equality of $$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) }.$$ </wikitex>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

$\displaystyle T'_{F(M),F(M)}(F (\varphi) \circ G(M))=e'_{F(M)}\circ T'G(M) \circ T'F(\varphi)$ $$\displaystyle T'_{F(M),F(M)}(F (\varphi) \circ G(M))=e'_{F(M)}\circ T'G(M) \circ T'F(\varphi)$$

and

$\displaystyle F(T_{M,M}\varphi)\circ G(M)=F(e_M)\circ FT\varphi\circ G(M).$ $$\displaystyle F(T_{M,M}\varphi)\circ G(M)=F(e_M)\circ FT\varphi\circ G(M).$$

This follows from the commutative diagram

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