Why is the coproduction in the tensor algebra a homomorphism? Let $V$ be a vector space over a field $K$. And let $T^k V = V \otimes V \otimes\ldots \otimes V $ ($k$-times). Then I am interested in the space
$$
T(V) = \bigoplus_{k=0}^\infty T^k V .
$$
The tensor product rule promotes $T(V)$ to a graded algebra known as the tensor algebra. We can also give a coalgebra structre to $T(V)$ by defining the coproduct as the map
$$ 
\Delta : T(V) \to T(V) \boxtimes T(V),
$$
where $\boxtimes$ denotes the exterior tensor product (just to distinguish it from $\otimes$). $\Delta$ takes the explicit form for $v \in T^1V =V$:
$$
\Delta : v \mapsto  1 \boxtimes v + v \boxtimes 1,
$$
and $\Delta:1 \mapsto 1 \boxtimes 1$. A central statement on the wiki page on the tensor algebra describes $\Delta$ as a homomorphism. And here I star to get puzzled. One extends the definition of $\Delta$, i.e.,
$$
\Delta ( v \otimes w ) =\Delta(v) \otimes \Delta(w),
$$
which clearly contains our notion of a homomorphism. However, I have the feeling that this is in conflict with our previous definition. If we expand the right hand side we find (as is also state on the wikipedia page from above)
$$ \Delta(v) \otimes \Delta(w) = \big[(v\otimes w) \boxtimes 1 + 1 \boxtimes (v \otimes w)\big] + v \boxtimes w + w \boxtimes v.$$
My confusion arises from the second term. I somehow want it to be zero such that only the term in brackets survives. I guess what I would like to see is that we can have a coproduct for $t \in T(V)$ which is defined as
$$
\Delta: t \mapsto 1 \boxtimes t + t \boxtimes 1 .
$$
Such a definition would also yield a coalgebra but $\Delta$ would not be an algebra homomorphism because of the "problematic term" from before. Is it possible to define a $\Delta$ in a similar way to my proposal but such that it is also an algebra homomorphism?
 A: You can get the general definition of $\Delta$ the same way you found $\Delta(v\otimes w)$.  An arbitrary element of $T(V)$ is a linear combination of elements of the form $v_1\otimes v_2\otimes\dots\otimes v_n$ where $v_i\in V$ for all $i$.  You then just define $$\Delta(v_1\otimes v_2\otimes\dots\otimes v_n)=\Delta(v_1)\otimes\Delta(v_2)\otimes\dots\otimes\Delta(v_n)$$ where $\Delta(v_i)=v_i\boxtimes 1+1\boxtimes v_i$ for each $i$.
Of course, it is not obvious that this is well-defined, since the representation of an element of $T(V)$ as a linear combination of elements of the form $v_1\otimes v_2\otimes\dots\otimes v_n$ is not unique.  There are a few ways you can prove that this really is well-defined.  One way is to pick a basis for $V$ and restrict the $v_i$ to be elements of the basis, which makes the representations unique.  Another way is to just invoke the universal property of $T(V)$: given any linear map $V\to A$ for some algebra $A$, it extends uniquely to an algebra homomorphism $T(V)\to A$.  Here we are taking $A=T(V)\boxtimes T(V)$ and our linear map is just $v\mapsto v\boxtimes 1+v\boxtimes 1$.
