Briefly, the recursion theorem states that all computable functions $f$ have a fixed point $n$ such that $\phi_n = \phi_{f(n)}$, where $\phi_e$ is a standard enumeration of partial computable functions.
Some extra knowledge is that Soare defines $d$ as the "diagonal" function, $d(u) = \phi_u(u)$. The proposition I am referring to states that the fixed point $n$ can be computed from $f$ by a $1:1$ computable function $g$, and offers as proof the following construction of $g$:
"Let $v(x)$ be a computable function such that $\phi_{v(x)} = \phi_x \circ d$. Let $g(x) = d(v(x))$. Both $d$ and $v$ are $1:1$ by the $s-m-n$ (or parameter) theorem."
When thinking about this proof I tried to write out $g(x)$ to see if I could get a better grip on what it did. Keeping in mind the definition of the diagonal function, we can expand the partial computable function with index $v(x)$ as $\phi_{v(x)}(y) = \phi_x(\phi_y(y))$. This being so we can now try to rewrite $g(x) = d(v(x))$ as $g(x) = \phi_{v(x)}(v(x)) = \phi_x(\phi_{v(x)}(v(x)))$. (!)
Herein lies my problem. What is going on? This function $g(x)$ seems to be impossible to actually pin down; it just keeps expanding. If anyone could help by pointing to where I have made a mistake, that would be great, thank you.
After some thought, I realized there was nothing wrong with this. This is simply an equational consequence of something being a fixed point. I suppose it was troubling briefly because I thought there ought to be some "resolution" to the function as you "expanded" by substitution, but really all this is saying is that a point is the same as a function applied once to the point, which is the same as the function applied twice to the point, etc.
So this makes sense of expressions like $n = f(n) = f(f(n)) = ... = f(...(f(n))...)$.
Hopefully this helps someone in the future who is also shocked by the recursion theorem at first glance.