Question about trivializing an SPT phase via group extension.

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Consider a d (spatial) dimensional SPT phase with an on-site symmetry G, classified by some non-trivial cocycle  $\omega^{d+1}(\{g_i\}) \neq \delta \mu^d(\{g_i\})$, $[\omega^{d+1}(\{g_i\})] \in H^{d+1}(G,U(1))$, . In a recent paper, Wang, Wen and Witten construct gapped boundaries via a suitable group extension $1 \longrightarrow H \overset{i}{\longrightarrow} K \overset{r}{\longrightarrow} G \longrightarrow 1$ such that the cocycle fro $H$ defined via pullback is trivial $r^*\omega^{d+1}(\{h_i\}) = \omega^{d+1}(\{r(h_i)\})= \delta \mu^d(\{h_i\})$. The gapped boundary corresponds to an $H$ invariant theory but with $K$ gauged so that the global symmetry is $H/K \cong G$ as required. The following sentence they say however confuses me:

" By definition, two states in two different $G$-SPT phases cannot smoothly deform into each other via deformation paths that preserve the $G$-symmetry. However, two such $G$-SPT states may be able to smoothly deform into each other if we view them as systems with the extended $H$-symmetry and deform them along the paths that preserve the $H$-symmetry. "

To me, it sounds like the two sentences contradict each other.
Q1) If there is an $H$ invariant deformation path to connect the system to a trivial state, then does that not automatically give us a $G$ invariant deformation path?
Q2) Does sentence 2 somehow only apply to the boundary rather than the bulk?

Please give a background reference. What is SPT?

What is SPT?

SPT phase = symmetry-protected topological phase

I apologize for the delayed response. The response by Xiao-Gang Wen in this post in physics stack exchange gives a nice explanation on what a Symmetry Protected Topological (SPT) phase is and is not. https://physics.stackexchange.com/questions/135398/definition-of-short-range-entanglement

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