# A question on the Robba ring

Notation is as in the question: https://math.stackexchange.com/questions/4090045/some-questions-about-the-robba-ring.

We define a new operator over the Robba ring as follows. Put $$c=\frac{pE(u)}{E(0)}$$ and define a Frobenius map $$\varphi$$ by $$\varphi(u)=u^p$$, then extend $$\varphi$$ semi-linearly to the Robba ring $$\mathcal{R}$$.

Now we have two operators over the Robba ring: $$c\varphi-1$$ and $$N_{\nabla}=-\lambda u\frac{d}{du}$$.

Question. Is the map $$(c\varphi-1) \oplus N_{\nabla}:\mathcal{R}\oplus \mathcal{R}\to \mathcal{R}$$ surjective?

Remarks:

1. I expect this is true. Some evidences are given in the remarks of the linked question above.
2. Besides the evidences in the link. It is also easy to see that $$\mathsf{Im}(c\varphi-1)\supset \mathcal{R}^{+}$$.
3. Counter-examples (series of $$\mathcal{R}$$ that are not in the image) are also welcomed, I tried but cannot find one neither.
4. The Appendix A.1 of Pierre Colmez's article "Représentations triangulines de dimension 2" is helpful but not enough to answer the question in my opinion.
5. Using facts from Galois cohomology, a different but related statement is true: $$(\varphi-1) \oplus (\gamma-1):\mathcal{R}\oplus \mathcal{R}\to \mathcal{R}$$ is surjective. Indeed, this follows from Theorem 2.8 of Ruochuan Liu's article "Cohomology and Duality for (phi,Gamma)-modules over the Robba ring". (One would expect $$N_{\nabla}$$ to be nicer than $$\gamma-1$$ to calculate with, while unfortunately the factor $$\lambda$$ in $$N_{\nabla}$$ makes it very complicated.)

The same question on StackExchange receives no answer yet, so I reput it here. Any answers, remarks or references are welcomed!