I am working with Friedland entropy and there is a proof I cannot figure out how to do. Friedland entropy is defined for $\mathbb{Z}^k$ continuos actions $\mathcal{T}$ on a topological metric space $X$. Let $T_i$ for $i=1,\dots,k$ be the generators of the actions, i.e. the images of the vectors of the canonical base $\mathcal{T}(e_i)=T_i$.

The orbit space is the defined $$ \mathcal{X}=\{(x_n)_{n \in \mathbb{N}}\in \prod_{n \in \mathbb{N}}X \ \ | \ \ \forall n \in \mathbb{N},\exists i=1\dots,k:T_i(x_n)=x_{n+1}\}. $$ On this space we have the shift transformation $\sigma$ defined with $\sigma((x_n)_{n \in \mathbb{N}})=(x_{n+1})_{n \in \mathbb{N}}$. We define the entropy of the action with $e(\mathcal{T})=h_{top}(\sigma)$: it is equal to the topological entropy of the shift.

I read in a paper by Pollicott ad Geller that if we have a $\mathbb{Z}^2$ action generated by a homeomorphism on a compact metric space $X$ and by the identity, then the Friedland entropy of the actions is equal to the topological entropuy of the homeomorphism. The article says it is trivial but I cannot figure out how to do that. Could you help me?

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