Q:

How to obtain $Z(\Sigma_f)=\text{Trace}\ \Sigma(f)$ in TQFT?

I am studying TQFT and have a question on one standard property. 
A remark in Wikipedia (see the link above) says:

If for a closed manifold $M$ we view $Z(M)$ as a numerical invariant, then for a manifold with boundary we should think of $Z(M) ∈ Z(∂M)$ as a "relative" invariant. Let $f : Σ × I → Σ × I$ be an orientation preserving diffeomorphism, and identify opposite ends of $Σ × I$ by $f$. This gives a manifold $Σ_f$ and our axioms imply
  $$
Z(\Sigma_f)=\text{Trace}\ \Sigma(f)
$$
  where $Σ(f)$ is the induced automorphism of $Z(Σ)$.

How can one see this formula? A standard way to cut the manifold is to cut it "at the middle" to obtain two manifolds with boundary, each of which is isomorphic to $Σ × I$, but this does not help me get the formula. 

A:

If you cut $\Sigma _f$ "in the middle", as you say, you obtain indeed two manifolds $M_1,M_2$ diffeomorphic to $\Sigma \times I$. For $M_1$ the two embeddings of $\Sigma $ in the boundary are standard, hence the element $Z(M_1)$ of $\mathrm{Hom(Z(\Sigma ),Z(\Sigma ))}$ is the identity.
For $M_2$ the two embeddings differ by $f$, hence  $Z(M_2)\in\mathrm{Hom(Z(\Sigma ),Z(\Sigma ))}$ is $\Sigma (f)$. 
Now the associativity axiom tell you that $Z(\Sigma_f )=\mathrm{Tr}(\mathrm{Id}\circ \Sigma (f))=\mathrm{Tr}(\Sigma (f))$.

