Why is the trace map the only evaluation map applicable to a TQFT Below is a quote from this paper found on page 8.

Dually, the evaluation map is ev(T) = tr(T) for some T ∈ hom(V, V ).

Why does $ev(T)$ have to be the trace of $T$?  It seems that there are a lot of maps from the vector space to the underlying field that we could use.  What conditions make it necessarily the trace map?
 A: Glad to see you're goin on studying TQFTs :)
The trace $ev$ map has a universal property; among many morphisms $V^\lor\otimes V\to k$ it's the unique map such that composed with a "dual" map $coev\colon k \to V \otimes V^\lor$ in suitable "zig-zags"
$$
\begin{array}{c}
V \xrightarrow{V\otimes ev} V\otimes V^\lor \otimes V \xrightarrow{coev \otimes V} V \\
V \xrightarrow{V\otimes coev} V^\lor \otimes V\otimes V^\lor \xrightarrow{ev\otimes V} V^\lor
\end{array}
$$
give the identity map. This definition works well in every category where there's a notion of "dual" of an object. It's a matter of a basic linear algebra course to see that if $ev$ is the trace of a linear operator and $coev$ is the map sending $1\in k$ into $\sum v_i\otimes v^i$ (subscript: elements of a basis of $V$; superscript: its dual basis) then you get precisely the zigzags identities.
There's a pretty much neat reason why this trace comes out, and that's the good monoidal structure of the category of vector spaces: the two-sided dual of $V$ is obtained as the internal hom $[V,k]$: then $ev$ and $coev$ are precisely the counit and unit, respectively, of the adjunction $-\otimes k\dashv \hom_k(-,k)$ (that's an autoequivalence of $\sf Vect$, of course), and zigzags identities are precisely those of the adjunction. 
Compact closed categories arise in several places, and homming with the unit object is a general technique as well.
An engineer (no flame intended) would say that sometimes you build duals by "twisting what seems to have period 2 and hope for best"; in vector spaces you transpose a matrix. In cobordisms you invert orientation. In spectra... well, it's more complicated there but the idea is almost the same.
Since a TQFT preserves duals (that's an instructive exercise), you get the idea that $Z(S^1\coprod \overline{S^1}\to \varnothing)$ is the trace map.
