quantum mechanicstopological-insulatorstopology
I know that a Dirac point carries a Chern number of $\pm\frac{1}{2}$ and when we have a gap closure at any point in our band structure we can transfer Chern numbers depending on at how many points we have gap closure. My question is, that if we start with a topologically trivial phase and close the gap and reopen it, are we guaranteed to have a topologically non-trivial phase?
For example, in HgTE quantum wells we get a non-trivial phase after band inversion.
I am still not sure what you precisely want to be a Klein Bottle, but let me make some comments that might help you clarify what exactly you want to know. (Warning: I am writing this while being very tired, people are invited to correct me.)
First of all one must be careful to distinguish band structure of the bulk from band structure of a semi-infinite strip (with edges). In your second figure, 1 og 2 are bulk band structures while 3 is band structure of a semi-infinite strip. I think you are mixing these.
This I mention, in case there were any misunderstanding here. Now to your question, how can one see the connection between band structure, Brillouin zone and topology? The interesting thing is that we only need to analyse the bulk in order to probe the topology of the system, although the interesting physics is along the edge (this is called bulk-edge correspondence in the litterature).
Assume we add a spin-orbit coupling to graphene that preserves $S_z$, meaning that although the spin $SU(2)$ symmetry is not preserved it is still well-defined to speak about spin-up/down (so a $U(1)$ subgroup is still a symmetry). I think that this term will open a gap at the Dirac points at $K$ and $K'$ and the system turns into an insulator. Since $K$ and $K'$ are not important anymore, we have a four-band model (two from spin up/down and two from sub lattice A/B). Since $S_z$ is conserved, the (bulk) Hamiltonian is block diagonal
$ H(\mathbf k) = \begin{pmatrix} H_{\uparrow}(\mathbf k) & \\ & H_{\downarrow}(\mathbf k) \end{pmatrix}$,
where $H_{\uparrow}$/$H_{\downarrow}$ are $2\times 2$ matrices. Since the system preserves time-reversal symmetry, the two Hamiltonians are related by a time-reversal transformation $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$. Since the Brillouin zone is a torus, we need to classify maps $T^2\rightarrow \mathcal H$, where $\mathcal H$ is the appropriate space of $4\times 4$ matrices obeying some constraints (gapped and time-reversal invariance). For this model, however, we don't need to analyze $\mathcal H$. Since the model is block diagonal we can consider each block separately, any $2\times 2$ matrix can be written in the basis of Pauli matrices
$H_{\alpha}(\mathbf k) = d_0(\mathbf k) \mathbf I + \mathbf d(\mathbf k)\cdot\mathbf{\sigma}$, where $\alpha = \uparrow, \downarrow$ (the dependence of $d_0$ and $\mathbf d$ on $\alpha$ is suppressed for notational simplicity).
Since the spectrum is $E(\mathbf k) = d_0(\mathbf k) + \sqrt{\mathbf d\cdot\mathbf d}$, we can continously deform $d_0$ to zero and deform $\mathbf d\rightarrow \hat{\mathbf d}=\frac{\mathbf d}{|\mathbf d|}$ without closing the gap (and thus still be in the same topological class). Thus we can classify each block of the Hamiltonian, by the map $T^2\rightarrow S^2$, $\mathbf k\rightarrow \hat{\mathbf d}(\mathbf k)$. This is basically classified by the second homotopy group of the 2-sphere $S^2$ (winding number), $\pi_2(S^2) = \mathbb Z$ (well the map is from a torus $T^2$ and not $S^2$, so we need an argument for why we can use the second homotopy group but this I will not delve into right now). The winding number of $\hat{\mathbf d}(\mathbf k)$ is given by the degree of map formula
$C_1^{\alpha} = \frac 1{4\pi}\int_{T^2}d\mathbf k\;\hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}\in\mathbb Z.$
Thus each block form a Integer Quantum Hall effect separately (see also this answer). Note that $C_1$ is basically the first Chern-number where $\epsilon^{\mu\nu}F_{\mu\nu} = \hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}$, $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ where $A_{\mu}(\mathbf k) = -i\langle\psi(\mathbf k)|\partial_{\mu}\psi(\mathbf k)\rangle$ is the Berry phase for the filled states (to see how to rewrite the Berry phase in terms of $\hat{\mathbf d}$ see this paper. The reason I have avoided the Chern number approach is that, one needs to go into discussing vector (or principal) bundles which just would complicate everything even more).
Now we are almost done. We have found two Chern numbers $C_1^{\downarrow}$ and $C_1^{\uparrow}$, is there a Chern number for the combined system? Since $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$ are related by time-reversal symmetry, one can show that $C_1^{\uparrow} = -C_1^{\downarrow}$. Thus the sum $C_1^{\uparrow} + C_1^{\downarrow} = 0$, but the difference $C_1^{\uparrow} - C_1^{\downarrow} = 2C_{spin}$ is well defined and sometimes called the spin Chern-number. Thus calculating $C_{spin}\in\mathbb Z$ will tell you if graphene with a spin-orbit coupling is topological or not.
However if one has a perturbation that breaks $S_z$, then our construction breaks down. But it turns out that $C_{spin}$ is still well defined but only modulo 2, $\nu = C_{spin} \text{mod} 2$. This is why people call this a $\mathbb Z_2$-topological insulator, see more details here (pdf-file).
There are of course many other ways of seeing the connection to topology and this is probably the most explicit and elementary way to do it. Is this what you wanted? I suspect that you are interested in the map $T^2\rightarrow\mathcal H$, $\mathbf k\rightarrow H(\mathbf k)$ and when you say "energy space" you mean $\mathcal H$. It turns out that this space is homotopy equivalent to $\mathcal H \approx U(8)/Sp(8)$ (called the symplectic class) by a band-flattening procedure. Sadly I am too tired to see if this is diffeomorphic to the Klein bottle. Is this what you are interested in?
This answer got much longer than originally planned.
Let me first answer your question "is it wrong to consider topological superconductors (such as certain p-wave superconductors) as SPT states? Aren't they actually SET states?"
(1) Topological superconductors, by definition, are free fermion states that have time-reversal symmetry but no U(1) symmetry (just like topological insulator always have time-reversal and U(1) symmetries by definition). Topological superconductor are not p+ip superconductors in 2+1D. But it can be p-wave superconductors in 1+1D.
(2) 1+1D topological superconductor is a SET state with a Majorana-zero-mode at the chain end. But time reversal symmetry is not important. Even if we break the time reversal symmetry, the Majorana-zero-mode still appear at chain end. In higher dimensions, topological superconductors have no topological order. So they cannot be SET states.
(3) In higher dimensions, topological superconductors are SPT states.
The terminology is very confusing in literature:
(1) Topological insulator has trivial topological order, while topological superconductors have topological order in 1+1D and no topological order in higher dimensions.
(2) 3+1D s-wave superconductors (or text-book s-wave superconductors which do not have dynamical U(1) gauge field) have no topological order, while 3+1D real-life s-wave superconductors with dynamical U(1) gauge field have a Z2 topological order. So 3+1D real-life topological superconductors (with dynamical U(1) gauge field and time reversal symmetry) are SET states.
(3) p+ip BCS superconductor in 2+1D (without dynamical U(1) gauge field) has a non-trivial topological order (ie LRE) as defined by local unitary (LU) transformations. Even nu=1 IQH state has a non-trivial topological order (LRE) as defined by LU transformations. Majorana chain is also LRE (ie topologically ordered). Kitaev does not use LU transformation to define LRE, which leads to different definition of LRE.
Best Answer
No. Take a free scalar theory $$ L=(d\phi)^2-m^2\phi^2 $$
If you scan $m\in(-\infty,+\infty)$, you go through the point $m=0$, where the gap closes. On either side, the theory is in the trivial topological phase.
You could do the same using free fermions, $$ L=\bar\psi d\psi-m\bar\psi\psi $$ As you scan $m\in(-\infty,+\infty)$, you again go through a gapless point. On either side, the theory may or may not be trivial. For example, in two dimensions, you get a non-trivial phase if and only if the number of fermion modes is odd. If you take an even number, then you are in the trivial phase on either side of $m=0$. In 3d you get a non-trivial phase for any number of fermions. In 4/5/6d you get a trivial phase regardless of the number of fermions. $d=7$ is similar to $d=3$. Etc.