Understanding cofree coalgebras 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?
 A: 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.
A: It's because $T(V)$ is graded.
Suppose that we decompose $g$ as $g = g_{0} \oplus g_{1} \oplus g_{2} \oplus \ldots$, where each $g_{i}$ is a function which maps only into the $i^\mathrm{th}$ graded piece of $T(V)$.
Then we can deduce $g_{0}$ from the counit law, $\mathrm{counit} \circ g = \mathrm{counit}$. So basically, $g_{0}$ is the counit in $C$, as a map into $0^\mathrm{th}$ graded piece of $T(V)$, which is isomorphic to $K$.
And we can deduce $g_{1} = f$ from $pg = f$.
Then, we can deduce $g_{2}, g_{3}, \ldots$ inductively from the comult law, $\mathrm{comult} \circ g = (g \otimes g) \circ \mathrm{comult}$. Specifically if $\mathrm{comult}_{i,n-i}$ means the part of comult that maps into the $(i,n-i)$ graded piece of $T(V)\otimes T(V)$, then we have $\mathrm{comult}_{i,n-i} \circ g_{n} = (g_{i} \otimes g_{n-i}) \circ \mathrm{comult}$, which enables $g_{n}$ to be determined (eg taking $i = 1$).
