I have troubles with understanding some points in Loday-Vallette book (Algebraic operads).
(1) Coderivations Let $T(V)$ be the tensor algebra over a vector space $V$. It is well known that it is a free algebra with obvious multiplication (=concatenation) $\mu:T(V) \otimes T(V) \to T(V)$, $$ \mu(v_1 \cdots v_p \otimes v_{p+1} \cdots v_{p+q}):=v_1 \cdots v_p v_{p+1} \cdots v_{p+q}, $$ and it is also a cofree conilpotent coalgebra (denoted by $T^c(V)$) with the comultiplication (=deconcatenation) $$ \Delta(v_1\cdots v_n):=\sum_{i=0}^n v_1 \cdots v_i \otimes v_{i+1} \cdots v_n. $$
By the Leibniz rule, any derivation on $T(V)$ is completely determined by its value on $v\in V$ (=generators), i.e., we have the bijection $$ \mathrm{Der}(T(V)) \simeq \mathrm{Hom}(V, T(V)), \qquad \{f:V \to T(V)\} \mapsto d_f, $$ where $d_f(v_1\cdots v_n): = \sum_{i=1}^n v_1 \cdots f(v_i) \cdots v_n.$
Next, coderivation is defined as follows. Let $C$ be a coalgebra, then a coderivation $c$ is a linear map $c:C \to C$ such that $\Delta \circ c = (\mathbf{id} \otimes c + c \otimes \mathbf{id}) \circ \Delta.$ And, as expected, the set of coderivation on the $T^c(V)$ can be also be defined in a ``dual'' fashion as derivation is. More formally we have a bijection $$ \mathrm{Coder}(T^c(V)) \simeq \mathrm{Hom}(T^c(V), V). $$
And this is trouble for me. I don't understand what is going on here. Does it mean that it is enough to define a linear map $c$ as follows $c(w) = \sum_{i=1}^\infty \alpha_i \mathbf{e}_i$ and thus $c$ is a coderivation? Here $\{\mathbf{e}_1, \ldots, \mathbf{e}_i, \ldots,\}$ is a basis of $V$ and almost all $\alpha_i(w)$ are zero. If so, how does co-Leibnitz rule show this?