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:*

- I expect this is true. Some evidences are given in the remarks of the linked question above.
- Besides the evidences in the link. It is also easy to see that $\mathsf{Im}(c\varphi-1)\supset \mathcal{R}^{+}$.
- Counter-examples (series of $\mathcal{R}$ that are not in the image) are also welcomed, I tried but cannot find one neither.
- 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.
- 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!