Sufficiency of relations in $\mathbf{2Cob}$ - about TQFTs I am in the midst of proving that 2D TQFTs are in one-to-one correspondence to commutative Frobenius algebras. 
A TQFT is a symmetric monoidal functor $Z$ from the category of cobordisms $\mathbf{2Cob}$ to the category of vector spaces over a field $\mathbb{k}$, $\mathbf{Vect}_{\mathbb{k}}$. 
I have looked at the sufficiency of a set of relations among the generators in $\mathbf{2Cob}$, and I wonder why it is important to have sufficiency of this set of relations. 
If you have a commutative Frobenius algebra and want to define a TQFT, where does it go wrong if we don't have a sufficient set of relations in $\mathbf{2Cob}$?
Thanks in advance!
 A: Let $C$ be an object of $2Cob$ made of one connected component, that is, $C$ is a circumference. Then as you know the objects $\emptyset$, $C$, $C\sqcup C$, $C\sqcup C\sqcup C$, ... form a skeleton of $2Cob$. I denote as $\mathcal{S}$ this skeleton. Moreover I denote as: $$S=\{ [1], [lE], [rE], [lF], [rF], [T] \}$$ the usual class of generators of $\mathcal{S}$. So $[1]$ is the cylinder $C \to C$, $[lE] \colon \emptyset \to C$, $[rE] \colon C \to \emptyset$, $[lF]\colon C \to C\sqcup C$, $[rF]\colon  C\sqcup C\to C$ and $[T]$ is the twisting map $C\sqcup C\to C\sqcup C$.
Finally, I denote as $R$ the sufficient class of the relations of $S$ for $\mathcal{S}$ that you are talking about in your question.
Now, let $\mathbb{K}$ be a field and let: $$(V,f\colon \mathbb{K} \to V,g \colon V \to \mathbb{K},h \colon V \to V\otimes V,i\colon V\otimes V\to V)$$ be a commutative Frobenius $\mathbb{K}$-algebra. Your question is: in order to define the corresponding 2-dimensional quantum field theory $F$, where do we need the sufficiency of the class $R$?
Well, at first one defines $F$ as the unique (if it exists) symmetric monoidal functor: $$(\mathcal{S},\sqcup,\emptyset,\tau)\to(\mathbb{K}\text{-Vect},\otimes,\mathbb{K},\sigma)$$ such that $FC=V$, $F[lE]=f$, $F[rE]=g$, $F[lF]=h$ and $F[rF]=i$. As the objects of $\mathcal{S}$ are the finite disjoint unions of copies of $C$ and $S$ is a generating class of the symmetric monoidal category $(\mathcal{S},\sqcup,\emptyset,\tau)$, a symmetric monoidal functor $F$ whose domain is $(\mathcal{S},\sqcup,\emptyset,\tau)$ is completely determined by the values that it assumes on the objects $\emptyset$, $C$ and the arrows $[1]$, $[T]$, $[lE]$, $[rE]$, $[lF]$, $[rF]$. However, if the functor $F$ is symmetric monoidal then the values on $\emptyset$, $[1]$ and $[T]$ are forced to be $\mathbb{K}$, $1_V$ and $\sigma_{(V,V)}$ respectively, where $\sigma_{(V,V)}$ is the twisting map $V\otimes V\to V\otimes V$. Hence, if $F$ exists then it is unique.
Now we need to verify that these assignments actually define a symmetric monoidal functor (and this is where the sufficiency of the class $R$ is needed). As I said before, since we want $F$ to be a symmetric monoidal functor, in the definition of $F$ we mean that $F\emptyset=\mathbb{K}$, $F[1]=1_V$ and $F[T]=\sigma_{(V,V)}$.
Observe that if $F$ preserves the compositions and converts the functor $\sqcup$ into the functor $\otimes$, it is the case that $F$ automatically preserves the identities, converts the $\emptyset$ into the $\mathbb{K}$ and sends $\tau$ to $\sigma$ (as it sends $[T]$ to $\sigma_{(V,V)}$). Therefore, in order to conclude, we only need to prove that $F$ preserves the composition and that it converts the functor $\sqcup$ into the functor $\otimes$. Now, as $R$ is a sufficient class of relations for $S$, every possible equality involving the elements of $S$, the composition and the $\sqcup$-relation only is a consequence of the relations of $R$. Hence, in order to prove that $F$ preserves the composition and converts $\sqcup$ into $\otimes$ (i.e. in order to conclude) it suffices to verify that this happens for the relations of $R$. But this is indeed the case, because $(V,f,g,h,i)$ is a commutative Frobenius $\mathbb{K}$-algebra, and so you know that the "images" through $F$ of the relations of $R$ are precisely the conditions that every commutative Frobenius $\mathbb{K}$-algebra verifies. Hence $F$ is a symmetric monoidal functor, i.e. a 2-dimensional topological quantum field theory.
Summing up, where do we need the sufficiency of $R$? We need it when we have to verify that, given a commutative Forbenius algebra, the corresponding 2-dimensional TQFT is actually a 2-dimensional TQFT, that is, a symmetric monoidal functor.
