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Let $V$ be a vector space over $K$ and $T(V)$ denote the tensor algebra on $V$. It is well known that $T(V)$ is the free algebra on $V$. I've been told that it is also the cofree coalgebra on $V$ with comultiplication $\Delta:T(V)\rightarrow T(V) \otimes T(V)$ given by $$\Delta(v_1\dots v_n)=\sum_{\sigma\in S_n}\sum_i v_{\sigma(1)}\dots v_{\sigma(i)}\otimes v_{\sigma(i+1)}\dots v_{\sigma(n)}$$ where juxtaposition is shorthand for "internal" tensors and $S_n$ is the symmetric group.

To prove $T(V)$ is the free algebra, one takes a $K$-algebra $A$ and linear map $f:V\rightarrow A$ and shows that there is a unique algebra map $g:T(V)\rightarrow A$ such that $g\iota=f$ where $\iota:V\rightarrow T(V)$ is the inclusion map. The map $g$ is defined by $g(v_1\dots v_n) = f(v_1)\dots f(v_n)$.

I would like to show that for each $K$-coalgebra $C$ and each linear map $f:C\rightarrow V$ there is a unique coalgebra map $g:C\rightarrow T(V)$ such that $pg=f$ where $p:T(V)\rightarrow V$ is the projection map. However, in the cofree case, I don't see how to define $g$ because there doesn't seem to be an analogous procedure to "distributing over tensors" for a coalgebra map. I can't understand why a coalgebra map $g$ satisfying $pg=f$ is uniquely determined by $f$.

Edit: now I'm not even sure that this comultiplication is coassociative. Is something wrong here?

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Are you sure that the symmetrizing is necessary? I think that for the cofree coalgebra you only need deconcatenation comultiplication. My answer assumes this anyway, so if that's not correct then nor is my answer. –  DavidA Jun 12 '12 at 20:55
On first glimpse, in this tiresome morning, I read "Understanding coffee coalgebras". –  Asaf Karagila Jul 14 '12 at 6:42
Wikipedia says this isn't the cofree coalgebra. See en.wikipedia.org/wiki/Cofree_coalgebra . –  Qiaochu Yuan May 24 '13 at 23:23
Yeah, I had the multiplication wrong. David A was right that. –  Seth Feb 4 '14 at 0:25

2 Answers 2

It's because T(V) is graded.

Suppose that we decompose $g$ as $g = g_{0} \oplus g_{1} \oplus g_{2} \oplus ...$, where each $g_{i}$ is a function which maps only into the i'th graded piece of T(V).

Then we can deduce $g_{0}$ from the counit law, $counit \circ g = counit$. So basically, $g_{0}$ is the counit in C, as a map into 0'th graded piece of T(V), which is isomorphic to K.

And we can deduce $g_{1} = f$ from $pg = f$.

Then, we can deduce $g_{2}, g_{3}, ...$ inductively from the comult law, $comult \circ g = (g \otimes g) \circ comult$. Specifically if $comult_{i,n-i}$ means the part of comult that maps into the (i,n-i) graded piece of $T(V)\otimes T(V)$, then we have $comult_{i,n-i} \circ g_{n} = (g_{i} \otimes g_{n-i}) \circ comult$, which enables $g_{n}$ to be determined (eg taking i = 1)

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Thank you for your response. However I'm not completely understanding what you are saying. Let $x\in A$. We define $g_i (x)$ as the part of $g(x)$ of degree $i$. It is clear that $g_1 (x) = f(x)$ because $pg=f$. Now I want to use $\Delta \circ g = (g\otimes g)\circ \Delta$ to find $g_2$, $g_3$, etc. Can anyone explain in detail how I would find, for example, $g_2$? I need to find everything that $\Delta \circ g = (g\otimes g)\circ \Delta$ forces to be in degree $2$ but I am not seeing how to do this. –  Seth Jun 12 '12 at 13:30
We can write the comult in $T(V)$ as a direct sum $\Delta = \Delta_{0,0} \oplus \Delta_{0,1} \oplus \Delta_{1,0} \oplus \Delta_{0,2} \oplus ...$. Then $(\Delta_{0,0} \oplus \Delta_{0,1} \oplus ...) \circ (g_{0} \oplus g_{1} \oplus ...) = ((g_{0}\otimes g_{0}) \oplus (g_{0}\otimes g_{1}) \oplus (g_{1}\otimes g_{0}) \oplus ...) \circ \Delta$. Then, equating pieces of equal "bi-grading", we get for example $\Delta_{1,1} \circ g_{2} = (g_{1} \otimes g_{1}) \circ \Delta$, unless I'm mistaken. –  DavidA Jun 12 '12 at 20:47
Just to clarify my previous comment a little, $\Delta_{i,j}$ would be the part of $\Delta$ in $T(V)$ which maps $V^{\otimes (i+j)}$ to $V^{\otimes i} \otimes V^{\otimes j}$, via deconcatenation. That is, $x_{1} \otimes ... \otimes x_{i} \otimes x_{i+1} \otimes ... \otimes x_{i+j} \mapsto (x_{1} \otimes ... \otimes x_{i}) \otimes (x_{i+1} \otimes ... \otimes x_{i+j})$ –  DavidA Jun 12 '12 at 21:15

it's been long time since you asked the question but, $(T(V),\Delta)$ as you described is not the cofree coalgebra, it is instead the conilpotent cofree coalgebra.

We will work in the non-unital case, so $\bar T(V) = \bigoplus_{n=1}^{\infty}V^{\otimes n}$ and $$\Delta(v_1\cdot \ldots \cdot v_n) = \sum_{i = 1}^{n-1}v_1\cdot \ldots \cdot v_i \otimes v_{i+1}\cdot \ldots \cdot v_n$$ with the convention that the empty sum is $0$.

It's not hard to show that this is coassociative and conilpotent, i.e. for all $v \in \bar T(V)$, there is a $n$ such that $\Delta^m(x) = 0$ for $m\ge n$. Here $\Delta^{n+1} = (\Delta \otimes id_{\bar T(V)^{\otimes n}}) \circ \Delta^n$ and of course $\Delta^0 = id_{\bar T(V)}$.

Then take a coassociative, conilpotent coalgebra $(C,\Delta_C)$ and a linear map $f:C \to V$ and define $g:C \to \bar T(V)$ as follows,

$$\forall v \in \bar T(V), g(v) = \sum_{n=0}^{\infty} \left( f^{\otimes n+1} \circ \Delta_C^{n}\right)(v)$$

This map is well defined because of the conilpotency assumption, there is only a finite number of terms that are not $0$. I let you check that it is indeed a coalgebra morphism, and $p \circ g = f$ is an easy fact. Unicity comes from the coassociativity, there might be many ways to define $\Delta^{n}$ but they all coincide.

The unital case is not that different, you also need a conilpotency assumption that is slightly different, you can check in "Algebraic Operads" by Loday-Vallette paragraph 1.2 for more on this.

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