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What's the difference between a formal linear sum and its actual value? And how do you know that $\epsilon_V$ sends the sum to its value? I worked out the definition, and I found that $\epsilon_V(\sum\lambda_v v)=\sum\lambda_v 1_{U(V)}(v)=\sum\lambda_vv$. How to tell from this equation that the sum on the RHS is not a formal sum but its value?
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Well, there is no way to tell the difference from notation. That would have to be implicit (unless you explicitly state via something like $v+_{FU} w$ refers to sums in $FU(V)$ and $v+_V w$ refers to sums in $V$.
Note that, for any sets, elements in $F(S)$ aren't subjects to any linear relations. So say, you start with $V=\mathbb{R}^2,$ then $FU(V)$ is the vector space whose elements are formal real linear combinations of elementso f $\mathbb{R}^2$ - but this has infinite dimension not two, since, for instance, the image of $(2,0)$ doesn't know that it ought to be twice $(1,0)$).
Accordingly, sending a formal sum to its value is really remembering what the linear relations are. For instance, in the above example, the formal sum $2(1,0)$ (in $FU(\mathbb{R}^2$, where it isn't equal to $(2,0)$) gets sent to $2(1,0)$ in $\mathbb{R}^2,$ where it is equal to $(2,0)$.
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One way to make the relationship concrete is to identify $S$ uniquely with a collection of functions into $\mathbb R$. That is, we define, for each $s\in S,$ the function $f_s:S\to \mathbb R:x\mapsto \delta_x^s$. Then, $FS:=\text{span}\{f_s:s\in S\}$ is a vector space over $\mathbb R$, and now the elements of $FV$, that is, the "formal sums" $do$ have meaning: $\sum_ic_if_{s_i}$ is a bona fide function $g:S\to \mathbb R.$
I have a blog post I've mostly written that I need to finish on this and other "formal" constructions. This is a topic I find many people only have a slippery grasp on and is somewhat hard to explain, even though it's a fairly simple idea, because people have had the habit of identifying expressions with their interpretations drilled into them throughout most of their mathematical careers even when inappropriate (e.g. for polynomials or logical formulas).
The definition of formal $k$-linear combinations on a set $S$ is syntactic expressions in a language consisting of $+$, $0$, $r\cdot$ for each $r\in k$, and $v_s$ for each $s\in S$. Let's say $S=\{x,y,z\}$ for concreteness. Then $(r_1\cdot v_x + 0) + (r_2\cdot v_y+r_3\cdot v_y)$ is an example formal $k$-linear combination assuming $r_1, r_2, r_3\in k$. The set of syntactic expressions you can make in a given language (usually specified by a signature) generated from a set (which effectively serve to label "variables") is called the term algebra. This is an inductively defined set, i.e. it's the smallest set closed under these syntactic operations. As such no term is infinite in size. The intuitive idea being capture is terms are the things we write on the page and those inherently can only be of finite size.
Often we want certain equations to hold, e.g. $t+0=t$, so we can add some "equations" to the signature (which are just pairs of terms) and then quotient the term algebra by the congruence generated by this set of pairs of terms. For formal $k$-linear combinations, the "equations" provided would be the axioms of a vector space, e.g. associativity of $+$ and $r_1\cdot v+r_2\cdot v=r'\cdot v$ where $r'=r_1+r_2$ where here the addition is the addition of the field $k$. The benefit of doing this, other than some convenience, is that it turns the term algebra into an actual vector space.
This process is the same process that is used for every free algebraic structure generated from a set, e.g. free commutative algebras aka polynomials.
Returning to formal $k$-linear combinations on the set $S$, once we've quotiented the term algebra, we can easily show that every equivalence class of terms is represented by a term of the form $r_{s_1}\cdot v_{s_1}+\cdots+r_{s_n}\cdot v_{s_n}$ where $s_1,\dots,s_n\in S$ and are distinct, and $r_{s_1},\dots,r_{s_n}\in k$. We write, $\sum_{s\in S'}r_s v_s$ with $S'$ a finite subset of $S$ as shorthand for the fully written out term. It's clear that all the relevant information is contained in a function $r_{(-)} : S'\to k$. Since $0\cdot v=0$ and $t+0=0$, we can extend this function to all of $S$ by assigning $0$ for $s\notin S'$. We can then identify each element of the quotiented term algebra with a function $r:S\to k$ that is $0$ except at finitely many points.
Of course, we can shortcut all this understanding and simply define formal $k$-linear combinations generated from $S$ as finitely supported $k$-valued functions on $S$. This is often convenient. However, if you don't know where this definition comes from, calling it "formal $k$-linear combinations" seems mysterious and it's not clear why we need finitely supported functions. The set of all functions $S\to k$ is also a perfectly fine vector space, it just isn't (isomorphic to) the free vector space on $S$ unless $S$ is finite.
As to $\varepsilon_V$, you know that the result is an element of $V$ and $\varepsilon_V$, being an arrow in $\mathbf{Vect}_k$ is a vector space homomorphism, so, by definition, it takes the addition (and other operations) of $FU(V)$, which is a "formal" addition, and maps it to the addition of $V$, which is a "real" addition. Of course, $V$ could itself be a vector space of formal linear combinations, but the point is that $\varepsilon_V$ has no choice but to interpret the "formal" operations in $FU(V)$ as the operation so $V$ whatever they are. Formal $k$-linear combinations are not, in general, elements of $V$.