3.2. The Monoidal Structure
We have to show that π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) bears a monoidal structure ( π΅ ( π ) , β π΅ β‘ ( π ) , I ) (\mathcal{Z}(\mathcal{C}),\otimes_{\mathcal{Z}(\mathcal{C})},I) , such that all the requirements for a monoidal category
given in Definition 4 are satisfied.
(Ad 4.1): The object I β π΅ β‘ ( π ) I\in\mathcal{Z}(\mathcal{C}) is
( I , 1 β , 1 1 ( β β β ) ) (I,1_{-},1_{1_{(-\otimes-)}}) .
The tensor product of objects:
(Ad 4.2): The tensor product of two objects
( A , R A , β , R ~ ( A | β , β ) ) β π΅ β‘ ( π ) ( B , R B , β , R ~ ( B | β , β ) ) (A,R_{A,-},\tilde{R}_{(A|-,-)})\otimes_{\mathcal{Z}(\mathcal{C})}(B,R_{B,-},\tilde{R}_{(B|-,-)})
is defined to be the triple
( A β B , ( R A β R B ) β , ( R ~ A β R ~ B ) ( β , β ) ) (A\otimes B,(R_{A}\otimes R_{B})_{-},(\tilde{R}_{A}\otimes\tilde{R}_{B})_{(-,-)}) , where:
(1)
The underlying π \mathcal{C} -object is the tensor product A β B A\otimes B in π \mathcal{C} .
(2)
By RemarkΒ (9 ), the underlying pseudonatural equivalence
( R A β R B ) β : ( A β B ) β β β β β ( A β B ) (R_{A}\otimes R_{B})_{-}:(A\otimes B)\otimes-\Rightarrow-\otimes(A\otimes B) assigns a 1-morphism ( R A β R B ) X (R_{A}\otimes R_{B})_{X}
to any object X β π X\in\mathcal{C} and a 2-morphism ( R A β R B ) f (R_{A}\otimes R_{B})_{f} to any
1-morphism f : X β Y f\colon X\to Y . These are given as follows:
( R A β R B ) X = ( A β R B , X ) β ( R A , X β B ) , (R_{A}\otimes R_{B})_{X}=(A\otimes R_{B,X})(R_{A,X}\otimes B),
OPEN ( R A β R B ) f = ( ( A β R B , f ) β ( R A , Y β B ) ) β
( A β R B , X ) β ( R A , f β B ) ) , (R_{A}\otimes R_{B})_{f}=((A\otimes R_{B,f})\circ(R_{A,Y}\otimes B))\cdot(A\otimes R_{B,X})\circ(R_{A,f}\otimes B)),
or in terms of a diagram:
A β B β X {\lx@inpgf@ignorespaces ABX} A β X β B {\lx@inpgf@ignorespaces AXB} X β A β B {\lx@inpgf@ignorespaces XAB} A β B β Y {\lx@inpgf@ignorespaces ABY} A β Y β B {\lx@inpgf@ignorespaces AYB} Y β A β B {\lx@inpgf@ignorespaces YAB} A β B β f \scriptstyle{\lx@inpgf@ignorespaces AB\otimes f} A β R B , X \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,X}} β A β R B , f {\lx@inpgf@ignorespaces\Uparrow A\otimes R_{B,f}} R A , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B} β R A , f β B {\lx@inpgf@ignorespaces\Uparrow R_{A,f}\otimes B} X β A β f \scriptstyle{\lx@inpgf@ignorespaces XA\otimes f} A β R B , Y \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,Y}} R A , Y β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,Y}\otimes B}
To show that these data constitute a pseudonatural equivalence,
we have to show that
( β β β β ) (\bullet\otimes{\to}\to) and ( β β β ) (\bullet\otimes{\Downarrow}) hold.
This can be done easily by pasting together the corresponding diagrams for
R A , β R_{A,-} and R B , β R_{B,-} .
(3)
The underlying modification
( R ~ A β R ~ B ) ( X , Y ) : ( A β R B , X β Y ) β ( R A , X β B β Y ) β ( X β A β R B , Y ) β ( X β R A , Y β B ) β ( A β R B , X β Y ) β ( R A , X β Y β B ) (\tilde{R}_{A}\otimes\tilde{R}_{B})_{(X,Y)}:(A\otimes R_{B,X}\otimes Y)(R_{A,X}\otimes B\otimes Y)(X\otimes A\otimes R_{B,Y})(X\otimes R_{A,Y}\otimes B)\Rightarrow(A\otimes R_{B,X\otimes Y})(R_{A,X\otimes Y}\otimes B)
is defined to be the pasting:
A β B β X β Y {\lx@inpgf@ignorespaces ABXY} A β X β Y β B {\lx@inpgf@ignorespaces AXYB} A β X β B β Y {\lx@inpgf@ignorespaces AXBY} ββββ
X β Y β A β B {\lx@inpgf@ignorespaces XYAB} X β A β Y β B {\lx@inpgf@ignorespaces XAYB} X β A β B β Y {\lx@inpgf@ignorespaces XABY} A β R B , X β Y \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,XY}} β A β R ~ ( B | X , Y ) {\lx@inpgf@ignorespaces\Uparrow A\otimes\tilde{R}_{(B|X,Y)}} R A , X β Y β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,XY}\otimes B} β β R A , X , R B , Y β 1 {\lx@inpgf@ignorespaces\Uparrow\otimes_{R_{A,X},R_{B,Y}}^{-1}} β R ~ ( A | X , Y ) β B {\lx@inpgf@ignorespaces\Uparrow\tilde{R}_{(A|X,Y)}\otimes B}
Again it is easy to verify that this satisfies
( β β ( β β β ) ) (\bullet\otimes({\to}\otimes\bullet)) and
( β β ( β β β ) ) (\bullet\otimes(\bullet\otimes{\to})) and hence is a modification.
To show that this definition gives in fact an object in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) , we have to
verify that
( β β ( β β β β β ) ) (\bullet\otimes(\bullet\otimes\bullet\otimes\bullet)) is satisfied.
The following picture shows the tetrahedron.
Those vertices in the picture that are vertices of the tetrahedron are
written in big capitals. The remaining vertices occur since they are needed
for the decomposition.
X β Y β Z β A β B {\lx@inpgf@ignorespaces XYZAB} ββ
A β X β Y β Z β B {\lx@inpgf@ignorespaces\scriptstyle{AXYZB}} X β Y β A β Z β B {\lx@inpgf@ignorespaces\scriptstyle{XYAZB}} ββ
X β A β Y β Z β B {\lx@inpgf@ignorespaces\scriptstyle{XAYZB}} A β B β X β Y β Z {\lx@inpgf@ignorespaces ABXYZ} A β X β Y β B β Z {\lx@inpgf@ignorespaces\scriptstyle{AXYBZ}} X β Y β A β B β Z {\lx@inpgf@ignorespaces XYABZ} A β X β B β Y β Z {\lx@inpgf@ignorespaces\scriptstyle{AXBYZ}} X β A β Y β B β Z {\lx@inpgf@ignorespaces\scriptstyle{XAYBZ}} X β A β B β Y β Z {\lx@inpgf@ignorespaces XABYZ}
The following picture gives a decomposition of the
tetrahedron into four smaller commutative diagrams.
A β X β Y β Z β B AXYZB X β A β Y β Z β B XAYZB X β Y β A β Z β B XYAZB X β Y β Z β A β B XYZAB A β B β X β Y β Z ABXYZ A β X β B β Y β Z AXBYZ A β X β Y β B β Z AXYBZ A β X β Y β Z β B AXYZB A β X β B β Y β Z AXBYZ X β A β B β Y β Z XABYZ A β X β Y β B β Z AXYBZ X β A β Y β B β Z XAYBZ A β X β Y β Z β B AXYZB X β A β Y β Z β B XAYZB A β X β Y β B β Z AXYBZ X β A β Y β B β Z XAYBZ A β X β Y β Z β B AXYZB X β A β Y β Z β B XAYZB X β Y β A β B β Z XYABZ X β Y β A β Z β B XYAZB
Two of them are tetrahedra of the form
( β β ( β β β β β ) ) (\bullet\otimes(\bullet\otimes\bullet\otimes\bullet)) ,
tensored by an object from the left and the right, respectively.
The upper of the two triangular prisms commutes by the axioms 4 . ( v β i β i ) 4.(vii)
with Ξ± = R ~ ( A | X , Y ) \alpha=\tilde{R}_{(A|X,Y)} and g = R Z , B g=R_{Z,B} ,
together with 4 . ( v β i β i β i ) 4.(viii) .
The lower commutes by 4 . ( v β i ) 4.(vi) , applied to Ξ² = R ~ ( B | Y , Z ) \beta=\tilde{R}_{(B|Y,Z)} and
f = R A , X f=R_{A,X} together with 4 . ( v β i β i β i ) 4.(viii) .
One can verify that this tensor product is in fact associative.
We shall often write ( A β B , R A β R B , R ~ A β R ~ B ) (A\otimes B,R_{A}\otimes R_{B},\tilde{R}_{A}\otimes\tilde{R}_{B})
as a shorthand symbol for the tensor product of objects in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) .
The tensor product of an object and a morphism:
(Ad 4.3):
Let ( f , R f , β ) : ( A , R A , β , R ~ ( A | β , β ) ) β ( A β² , R A β² , β , R ~ ( A β² | β , β ) ) (f,R_{f,-})\colon(A,R_{A,-},\tilde{R}_{(A|-,-)})\to(A^{\prime},R_{A^{\prime},-},\tilde{R}_{(A^{\prime}|-,-)})
be a morphism in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) and let ( B , R B , β , R ~ ( B | β , β ) ) (B,R_{B,-},\tilde{R}_{(B|-,-)}) be an object.
Their tensor product is the morphism given by the pair
( f β B , ( β ( f , R B , β ) β ( R A β² , β β B ) ) β
( ( A β R B , β ) β ( R f , β β B ) ) ) , (f\otimes B,(\otimes_{(f,R_{B,-})}\circ(R_{A^{\prime},-}\otimes B))\cdot((A\otimes R_{B,-})\circ(R_{f,-}\otimes B))),
or in terms of a diagram:
A β B β X {\lx@inpgf@ignorespaces ABX} A β X β B {\lx@inpgf@ignorespaces AXB} X β A β B {\lx@inpgf@ignorespaces XAB} A β² β B β X {\lx@inpgf@ignorespaces A^{\prime}BX} A β² β X β B {\lx@inpgf@ignorespaces A^{\prime}XB} X β A β² β B {\lx@inpgf@ignorespaces XA^{\prime}B} f β B β X \scriptstyle{\lx@inpgf@ignorespaces f\otimes BX} A β R B , X \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,X}} β β f , R B , X {\lx@inpgf@ignorespaces\Uparrow\otimes_{f,R_{B,X}}} R A , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B} β R f , X β B {\lx@inpgf@ignorespaces\Uparrow R_{f,X}\otimes B} X β f β B \scriptstyle{\lx@inpgf@ignorespaces X\otimes f\otimes B} A β² β R B , X \scriptstyle{\lx@inpgf@ignorespaces A^{\prime}\otimes R_{B,X}} R A β² , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A^{\prime},X}\otimes B}
(Ad 4.4)
Let ( f , R f , β ) : ( B , R B , β , R ~ ( B | β , β ) ) β ( B β² , R B β² , β , R ~ ( B β² | β , β ) ) (f,R_{f,-}):(B,R_{B,-},\tilde{R}_{(B|-,-)})\to(B^{\prime},R_{B^{\prime},-},\tilde{R}_{(B^{\prime}|-,-)})
be a morphism in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) and let ( A , R A , β , R ~ ( A | β , β ) ) (A,R_{A,-},\tilde{R}_{(A|-,-)})
be an object. Their tensor product is the pair
( A β f , ( ( A β R f , β ) β ( R A , β β B β² ) ) β
( ( A β R B , β ) β β R A , β , f ) ) , (A\otimes f,((A\otimes R_{f,-})\circ(R_{A,-}\otimes B^{\prime}))\cdot((A\otimes R_{B,-})\circ\otimes_{R_{A,-},f})),
or in terms of a diagram:
A β B β X {\lx@inpgf@ignorespaces ABX} A β X β B {\lx@inpgf@ignorespaces AXB} X β A β B {\lx@inpgf@ignorespaces XAB} A β B β² β X {\lx@inpgf@ignorespaces AB^{\prime}X} A β X β B β² {\lx@inpgf@ignorespaces AXB^{\prime}} X β A β B β² {\lx@inpgf@ignorespaces XAB^{\prime}} A β f β X \scriptstyle{\lx@inpgf@ignorespaces A\otimes f\otimes X} A β R B , X \scriptstyle{\lx@inpgf@ignorespaces AR_{B,X}} β A β R f , X {\lx@inpgf@ignorespaces\Uparrow A\otimes R_{f,X}} R A , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}B} β β R A , X , f {\lx@inpgf@ignorespaces\Uparrow\otimes_{R_{A,X},f}} X β A β f \scriptstyle{\lx@inpgf@ignorespaces XA\otimes f} A β R B β² , X \scriptstyle{\lx@inpgf@ignorespaces AR_{B^{\prime},X}} R A , X β B β² \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}B^{\prime}}
To verify that these formulas really define morphisms in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) ,
one must check that ( β β β ) ({\to}\otimes{\to}) and
( β β ( β β β ) ) ({\to}\otimes(\bullet\otimes\bullet)) hold.
We only do this for 4.4 4.4 ; the other case being
similar. To show ( β β β ) ({\to}\otimes{\to}) one pastes together two cubes,
one being the ( β β β ) ({\to}\otimes{\to}) cube for f : B β B β² f\colon B\to B^{\prime} and
g : X β Y g\colon X\to Y
tensored on the left by A A , the other being a special case of 5 . ( v β i β i ) 5.(vii) .
For ( β β ( β β β ) ) ({\to}\otimes(\bullet\otimes\bullet))
we must show the following diagram commutes:
A β B β X β Y {\lx@inpgf@ignorespaces ABXY} A β X β Y β B {\lx@inpgf@ignorespaces AXYB} A β X β B β Y {\lx@inpgf@ignorespaces AXBY} ββββ
X β Y β A β B {\lx@inpgf@ignorespaces XYAB} X β A β Y β B {\lx@inpgf@ignorespaces XAYB} A β B β² β X β Y {\lx@inpgf@ignorespaces AB^{\prime}XY} X β A β B β Y {\lx@inpgf@ignorespaces XABY} A β X β Y β B β² {\lx@inpgf@ignorespaces AXYB^{\prime}} A β X β B β² β Y {\lx@inpgf@ignorespaces AXB^{\prime}Y} X β Y β A β B β² {\lx@inpgf@ignorespaces XYAB^{\prime}} X β A β Y β B β² {\lx@inpgf@ignorespaces XAYB^{\prime}} X β A β B β² β Y {\lx@inpgf@ignorespaces XAB^{\prime}Y} A β R B , X β Y \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,XY}} R A , B β X β Y \scriptstyle{\lx@inpgf@ignorespaces R_{A,B}\otimes XY} β A β R ~ ( B | X , Y ) {\lx@inpgf@ignorespaces\Uparrow A\otimes\tilde{R}_{(B|X,Y)}} 1 . {\lx@inpgf@ignorespaces 1.} R A , X β Y β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,XY}\otimes B} 7 . {\lx@inpgf@ignorespaces 7.} 8 . {\lx@inpgf@ignorespaces 8.} β β R A , X , R B , Y β 1 {\lx@inpgf@ignorespaces\Uparrow\otimes_{R_{A,X},R_{B,Y}}^{-1}} 2 . {\lx@inpgf@ignorespaces 2.} X β Y β R A , B \scriptstyle{\lx@inpgf@ignorespaces XY\otimes R_{A,B}} β R ~ ( A | X , Y ) β B {\lx@inpgf@ignorespaces\Uparrow\tilde{R}_{(A|X,Y)}\otimes B} 4 . {\lx@inpgf@ignorespaces 4.} A β R B β² , X β Y \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B^{\prime},X}\otimes Y} 5 . {\lx@inpgf@ignorespaces 5.} R A , X β B β² β Y \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B^{\prime}Y} 6 . {\lx@inpgf@ignorespaces 6.} X β R A , Y β B β² \scriptstyle{\lx@inpgf@ignorespaces X\otimes R_{A,Y}\otimes B^{\prime}} X β A β R B β² , Y \scriptstyle{\lx@inpgf@ignorespaces XA\otimes R_{B^{\prime},Y}} 3 . {\lx@inpgf@ignorespaces 3.}
1 . = A β R f , X β Y 2 . = β R A , X , f β Y 3 . = X β A β R f , Y 4 . = X β β R A , Y , f 5 . = A β R f , X β Y 6 . = β R A , X β Y , f 7 . = A β X β R f , Y 8 . = β R A , X , Y β f \begin{array}[]{llll}1.\>=\>A\otimes R_{f,X}\otimes Y&2.\>=\>\otimes_{R_{A,X},f\otimes Y}&3.\>=\>X\otimes A\otimes R_{f,Y}&4.\>=\>X\otimes\otimes_{R_{A,Y},f}\\
5.\>=\>A\otimes R_{f,X\otimes Y}&6.\>=\>\otimes_{R_{A,X\otimes Y},f}&7.\>=\>A\otimes X\otimes R_{f,Y}&8.\>=\>\otimes_{R_{A,X},Y\otimes f}\end{array}
We cut it into one rectangular and two triangular prisms.
To see that the left triangular prism commutes, we apply
( β β ( β β β ) ) ({\to}\otimes(\bullet\otimes\bullet)) to ( f , R f , β ) (f,R_{f,-}) ,
tensored on the left by A A .
The rectangular prism commmutes by 5 . ( v β i ) 5.(vi) and 5 . ( v β i β i β i ) 5.(viii) , applied to the
2-morphism R f , Y R_{f,Y} .
The right triangular prism commutes by 5 . ( i β v ) 5.(iv) and 5 . ( v β i β i ) 5.(vii) , applied to the
2-morphism R ~ ( A | X , Y ) \tilde{R}_{(A|X,Y)} .
The tensor product of an object and a 2-morphism:
(Ad 4.5):
For any object ( A , R A , β , R ~ ( A | β , β ) ) (A,R_{A,-},\tilde{R}_{(A|-,-)}) and any
2-morphism
Ξ± : ( f , R f , β ) β ( f β² , R f β² , β ) \alpha\colon(f,R_{f,-})\Rightarrow(f^{\prime},R_{f^{\prime},-}) we have a 2-morphism
A β Ξ± : ( A β f , β¦ ) β ( A β f β² , β¦ ) A\otimes\alpha:(A\otimes f,\dots)\Rightarrow(A\otimes f^{\prime},\dots)
(Ad 4.6):
For any object ( B , R B , β , R ~ ( B | β , β ) ) (B,R_{B,-},\tilde{R}_{(B|-,-)}) and any 2-morphism
Ξ± : ( g , R g , β ) β ( g β² , R g β² , β ) \alpha:(g,R_{g,-})\Rightarrow(g^{\prime},R_{g^{\prime},-}) we have a 2-morphism
Ξ± β B : ( g β B , β¦ ) β ( g β² β B , β¦ ) \alpha\otimes B:(g\otimes B,\dots)\Rightarrow(g^{\prime}\otimes B,\dots)
We must verify that these are 2-morphisms in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) , so we must check
( β β β ) ({\Downarrow}\otimes\bullet) .
We do this only for 4.5 4.5 .
A β B β X {\lx@inpgf@ignorespaces ABX} β A β Ξ± β X {\lx@inpgf@ignorespaces\Downarrow A\otimes\alpha\otimes X} A β B β² β X {\lx@inpgf@ignorespaces AB^{\prime}X} A β X β B {\lx@inpgf@ignorespaces AXB} β A β X β Ξ± {\lx@inpgf@ignorespaces\Downarrow AX\otimes\alpha} A β X β B β² {\lx@inpgf@ignorespaces AXB^{\prime}} X β A β B {\lx@inpgf@ignorespaces XAB} β X β A β Ξ± {\lx@inpgf@ignorespaces\Downarrow XA\otimes\alpha} X β A β B β² {\lx@inpgf@ignorespaces XAB^{\prime}} 2 . {\lx@inpgf@ignorespaces 2.} A β R B , X \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,X}} A β R B β² , X \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B^{\prime},X}} R A , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B} 4 . {\lx@inpgf@ignorespaces 4.} R A , X β B β² \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B^{\prime}} 1 . {\lx@inpgf@ignorespaces 1.} 3 . {\lx@inpgf@ignorespaces 3.}
1 . = A β R f , X 2 . = A β R g , X 3 . = β R A , X , f 4 . = β R A , X , g \begin{array}[]{ll}1.\>=\>A\otimes R_{f,X}&2.\>=\>A\otimes R_{g,X}\\
3.\>=\>\otimes_{R_{A,X},f}&4.\>=\>\otimes_{R_{A,X},g}\end{array}
The upper prism commutes by ( β β β ) ({\Downarrow}\otimes\bullet) tensored from the left
by A A . The lower prism commutes by an application of the axiom 4 . ( v β i ) 4.(vi) for
monoidal 2 2 -categories to the 2-morphism Ξ± \alpha .
The tensor product of morphisms:
(Ad 4.7):
For any morphisms
( f , R f , β ) : ( A , R A , R ~ A ) β ( A β² , R A β² , R ~ A β² ) (f,R_{f,-}):(A,R_{A},\tilde{R}_{A})\to(A^{\prime},R_{A^{\prime}},\tilde{R}_{A^{\prime}}) and
( g , R g , β ) : ( B , R B , R ~ B ) β ( B β² , R B β² , R ~ B β² ) (g,R_{g,-}):(B,R_{B},\tilde{R}_{B})\to(B^{\prime},R_{B^{\prime}},\tilde{R}_{B^{\prime}})
we have a 2-isomorphism:
β ( f , R f , β ) , ( g , R g , β ) := β f , g \otimes_{(f,R_{f,-}),(g,R_{g,-})}:=\otimes_{f,g}
To verify that this is a 2-morphism in π΅ β‘ ( π ) \mathcal{Z}(\mathcal{C}) , we have to check
( β β β ) ({\Downarrow}\otimes\bullet) .
The following diagram gives the proof.
A β B β X {\lx@inpgf@ignorespaces ABX} ββββ A β² β B β X {\lx@inpgf@ignorespaces A^{\prime}BX} A β B β² β X {\lx@inpgf@ignorespaces AB^{\prime}X} A β² β B β² β X {\lx@inpgf@ignorespaces A^{\prime}B^{\prime}X} A β X β B {\lx@inpgf@ignorespaces AXB} A β² β X β B {\lx@inpgf@ignorespaces A^{\prime}XB} A β X β B β² {\lx@inpgf@ignorespaces AXB^{\prime}} A β² β X β B β² {\lx@inpgf@ignorespaces A^{\prime}XB^{\prime}} X β A β B {\lx@inpgf@ignorespaces XAB} X β A β² β B {\lx@inpgf@ignorespaces XA^{\prime}B} X β A β B β² {\lx@inpgf@ignorespaces XAB^{\prime}} X β A β² β B β² {\lx@inpgf@ignorespaces XA^{\prime}B^{\prime}} A β R B , X \scriptstyle{\lx@inpgf@ignorespaces A\otimes R_{B,X}} f β B β X \scriptstyle{\lx@inpgf@ignorespaces f\otimes BX} β β f , g β X {\lx@inpgf@ignorespaces\Uparrow\otimes_{f,g}\otimes X} 1 . {\lx@inpgf@ignorespaces 1.} A β² β g Γ X \scriptstyle{\lx@inpgf@ignorespaces A^{\prime}\otimes g\times X} 3 . {\lx@inpgf@ignorespaces 3.} f β B β² β X \scriptstyle{\lx@inpgf@ignorespaces f\otimes B^{\prime}X} 2 . {\lx@inpgf@ignorespaces 2.} A β² β R B β² , X \scriptstyle{\lx@inpgf@ignorespaces A^{\prime}\otimes R_{B^{\prime},X}} R A , X β B \scriptstyle{\lx@inpgf@ignorespaces R_{A,X}\otimes B} 5 . {\lx@inpgf@ignorespaces 5.} β β f β X , g {\lx@inpgf@ignorespaces\Uparrow\otimes_{f\otimes X,g}} 4 . {\lx@inpgf@ignorespaces 4.} 7 . {\lx@inpgf@ignorespaces 7.} f β X β B β² \scriptstyle{\lx@inpgf@ignorespaces f\otimes XB^{\prime}} 6 . {\lx@inpgf@ignorespaces 6.} R A β² , X β B β² \scriptstyle{\lx@inpgf@ignorespaces R_{A^{\prime},X}\otimes B^{\prime}} X β A β g \scriptstyle{\lx@inpgf@ignorespaces XA\otimes g} 8 . {\lx@inpgf@ignorespaces 8.} X β f β B β² \scriptstyle{\lx@inpgf@ignorespaces X\otimes f\otimes B^{\prime}} β X β β f , g {\lx@inpgf@ignorespaces\Uparrow X\otimes\otimes_{f,g}}
1 . = A β R g , X 2 . = β f , R B β² , X 3 . = A β² β R g , X 4 . = β f , R B , X 5 . = β R A , X , g 6 . = R f , X β B β² 7 . = β R A β² , X , g 8 . = R f , X β B \begin{array}[]{llll}1.\>=\>A\otimes R_{g,X}&2.\>=\>\otimes_{f,R_{B^{\prime},X}}&3.\>=\>A^{\prime}\otimes R_{g,X}&4.\>=\>\otimes_{f,R_{B,X}}\\
5.\>=\>\otimes_{R_{A,X},g}&6.\>=\>R_{f,X}\otimes B^{\prime}&7.\>=\>\otimes_{R_{A^{\prime},X},g}&8.\>=\>R_{f,X}\otimes B\end{array}
The top cube commutes by 4 . ( i β v ) , ( v β i ) , ( v β i β i β i ) 4.(iv),(vi),(viii) , applied to the 2-morphism
A β R g , X A\otimes R_{g,X} .
The bottom cube commutes by 4 . ( i β v ) , ( v β i β i ) , ( v β i β i β i ) 4.(iv),(vii),(viii) , applied to the 2-morphism
R f , X β B R_{f,X}\otimes B .
We have to verify that these data satisfy the conditions 4 . ( i ) β ( v β i β i β i ) 4.(i)-(viii) .
These follow from the corresponding conditions holding in π \mathcal{C} .