# Why is the functor of tensor coalgebra right adjoint to the forgetful functor?

Let $\mathbf{NCoalg}$ be the category of non-unital coassociative conilpotent coalgebras (where "conilpotent" means that for any element there exists a power of comultiplication that vanishes on this element). We have a forgetful functor $\mathbf{NCoalg} \to \mathbf{Vect}$. Its seems that the right adjoint has to be the functor of non-unital tensor coalgebra: $$T(V)=V\oplus V^{\otimes 2}\oplus V^{\otimes3}\oplus \cdots ,$$ $$\Delta(v_1\otimes v_2\otimes \cdots \otimes v_n) = (v_1)\otimes(v_2\otimes\cdots\otimes v_n)+\cdots+(v_1 \otimes\cdots\otimes v_{n-1})\otimes(v_n).$$

Indeed, let $C$ be a coalgebra and $V$ a vector space. Consider a linear map $f: C \to V$. We would like to define a coalgebra map $\tilde{f}:C\to T(V)$ as follows: $$c \mapsto f(c)+f^{\otimes2}\Delta(c)+f^{\otimes3}\Delta^2(c)+\cdots.$$

The sum is in fact finite because $C$ is conilpotent.

However, how can I show that $\tilde{f}^{\otimes2}\Delta(c)=\Delta(\tilde{f}(c))$? I guess it can be done with some abominable coordinates but they appeared too abominable and I couldn't break through. Is there a better way (more categorical perhaps)?

• If $v\in V\subset T(V)$, what would $\Delta(v)$ be according to this formula ? Is it zero ? – Captain Lama Apr 18 '16 at 11:02
• @CaptainLama Yes, it is zero ($v$ is primitive). – Najib Idrissi Apr 18 '16 at 11:05

It's actually not that hard, but it needs a little care (for regular people like me who find working with coalgebras counter-intuitive :) ).

It all boils down to the fact that if $\Delta(c) = \sum_i x_i\otimes y_i$ then $$\Delta\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) = \sum_i \sum_{p+q=n} \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left(f^{\otimes (q+1)}\Delta^q(y_i)\right).$$

Indeed, coassociativity tells you that in $C^{\otimes (n+2)}$, $\Delta^{n+1}(c) = \sum_i \Delta^p(x_i)\otimes \Delta^q(y_i)$ for any $p,q$ such that $p+q = n$, when you identify $C^{\otimes (n+2)} \simeq C^{\otimes (p+1)}\otimes C^{\otimes (q+1)}$.

So if we call $\Phi_{p,q}: V^{\otimes (n+2)} \to V^{\otimes (p+1)}\otimes V^{\otimes (q+1)}$ the canonical isomorphism you get $\Phi_{p,q}\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) = \sum_i \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left( f^{\otimes (q+1)}\Delta^q(y_i)\right)$.

But now if $x\in V^{\otimes (n+2)}$ then $\Delta(x) = \sum_{p+q=n} \Phi_{p,q}(x)$ by definition, so if you apply $f^{\otimes(n+2)}$ and $\Delta$ to the formula above expressing the coassociativity, you get the formula emphasized at the beginning.

Now we just need to unravel definitions :

$$\begin{eqnarray*} \widetilde{f}^{\otimes 2}\Delta(c) & = & \sum_i \widetilde{f}(x_i)\otimes \widetilde{f}(y_i) \\ & = & \sum_i \left( \sum_{p\geqslant 0} f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left( \sum_{q\geqslant 0} f^{\otimes (q+1)}\Delta^q(y_i)\right) \\ & = & \sum_{n\geqslant 0} \sum_i \sum_{p+q=n} \left(f^{\otimes (p+1)}\Delta^p(x_i)\right)\otimes \left(f^{\otimes (q+1)}\Delta^q(y_i)\right) \\ & = & \sum_{n\geqslant 0} \Delta\left(f^{\otimes (n+2)}\Delta^{n+1}(c)\right) \\ & = & \sum_{n\geqslant 0} \Delta\left(f^{\otimes (n+1)}\Delta^n(c)\right) \\ & = & \Delta(\widetilde{f}(c)). \end{eqnarray*}$$

• Thanks a lot! I'm indeed a very regular person who finds multiplication a lot handier than comultiplication. But yes, your explanation was very clear. – Polydarya Apr 18 '16 at 13:06