My problem concerns with the unique solvability of a linear system of integral equations. In my problem, as I was able to write the system in matrix form: $$ M \begin{align} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} f \\ g \end{bmatrix} \end{align} $$ where $$ M = \begin{align} \begin{bmatrix} I - K_1 & S_1 \\ K_2 & S_2 \end{bmatrix} \end{align} $$ is an operator define on $H^{\frac12}(\Gamma_1)\times H^{\frac12}(\Gamma_2)$ ($\Gamma_1 \cap \Gamma_2 = \emptyset$, $\Gamma_1, \Gamma_2$ have certain smoothness).

To show the well-posedness of the above equation, I just need to show that $M$ is invertible. I doing so, I followed the ideas in Chapter 4-4 of Taylor's book on Pseudo Differential Operators. I understand that the steps are as follows:

1. Show that the adjoint $M^*$ of $M$ is injective.
2. Show that $M^*$ has a closed-range.
3. Show that $M$ is an isomorphism.

I was able to do the first step but I am stuck in showing the last two.

Here are my questions.

First Question I am not sure, but, do I really need to do the second step? I think that if could show that $M$ is an isomorphism, then the condition that the range of $M^*$ is closed is already satisfied (kindly correct me if my argument is wrong and please point out what might be my mistake).

Second Question Now, I believe I can apply Fredholm alternative to show that $M$ is in fact bijective. Knowing that, I need to show that I can express $M$ as a sum of the identity operator and a compact operator, i.e., $M=I+C$, where $C$ is a compact operator. For this one, I believe I can use the condition that $K_1, K_2, S_1, S_2$ are compact, but, I don't know how to do this. Is my argument correct? and how to do I show that $M$ is a sum of the identity operator and a compact operator?

Thank you.