Editing Tensor algebra (section)

===Counit===
The counit <math>\epsilon : TV \to K</math> is given by the projection of the field component out from the algebra. This can be written as <math>\epsilon: v\mapsto 0 </math> for <math>v\in V</math> and <math>\epsilon: k\mapsto k </math> for <math>k\in K=T^0V</math>.  By homomorphism under the tensor product <math>\otimes</math>, this extends to 
:<math>\epsilon: x\mapsto 0 </math>
for all <math>x\in T^1V \oplus T^2V\oplus \cdots</math>
It is a straightforward matter to verify that this counit satisfies the needed axiom for the coalgebra:
:<math>(\mathrm{id} \boxtimes \epsilon) \circ \Delta = \mathrm{id} = (\epsilon \boxtimes \mathrm{id}) \circ \Delta.</math>
Working this explicitly, one has
:<math>\begin{align}
((\mathrm{id} \boxtimes \epsilon) \circ \Delta)(x)
&=(\mathrm{id} \boxtimes \epsilon)(1\boxtimes x + x \boxtimes 1) \\
&=1\boxtimes  \epsilon(x) + x \boxtimes  \epsilon(1) \\
&=0 + x \boxtimes 1 \\
&\cong x
\end{align}</math>
where, for the last step, one has made use of the isomorphism <math>TV\boxtimes K \cong TV</math>, as is appropriate for the defining axiom of the counit.