Editing Tensor algebra (section)

=== Compatibility ===
The unit and counit, and multiplication and comultiplication, all have to satisfy compatibility conditions. It is straightforward to see that 
:<math>\epsilon \circ \eta = \mathrm{id}_K.</math>
Similarly, the unit is compatible with comultiplication:
:<math>\Delta \circ \eta = \eta \boxtimes \eta \cong \eta</math>
The above requires the use of the isomorphism <math>K\boxtimes K \cong K</math> in order to work; without this, one loses linearity. Component-wise,
:<math>(\Delta \circ \eta)(k) = \Delta(k) = k(1 \boxtimes 1) \cong k </math>
with the right-hand side making use of the isomorphism.

Multiplication and the counit are compatible:
:<math>(\epsilon \circ \nabla)(x\boxtimes y) = \epsilon(x\otimes y) = 0</math>
whenever ''x'' or ''y'' are not elements of <math>K</math>, and otherwise, one has scalar multiplication on the field: <math>k_1\otimes k_2=k_1 k_2.</math>  The most difficult to verify is the compatibility of multiplication and comultiplication:
:<math>\Delta \circ\nabla = (\nabla \boxtimes \nabla) 
\circ (\mathrm{id} \boxtimes \tau \boxtimes \mathrm{id}) 
\circ (\Delta \boxtimes \Delta)</math>
where <math>\tau(x\boxtimes y)= y \boxtimes x</math> exchanges elements. The compatibility condition only needs to be verified on <math>V\subset TV</math>; the full compatibility follows as a homomorphic extension to all of <math>TV.</math> The verification is verbose but straightforward; it is not given here, except for the final result:
:<math>(\Delta \circ\nabla)(v\boxtimes w) = \Delta(v\otimes w)</math>
For <math>v,w\in V,</math> an explicit expression for this was given in the coalgebra section, above.