I am studying [TQFT](http://en.wikipedia.org/wiki/Topological_quantum_field_theory) 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. 



How to obtain $Z(\Sigma_f)=\text{Trace}\ \Sigma(f)$ in TQFT?
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 $Z(\Sigma\_f )=\mathrm{Tr}(\mathrm{Id}\circ \Sigma (f))=\mathrm{Tr}(\Sigma (f))$.


