Let $X$ be a Banach space, $Z \subset X$ a closed, linear subspace and $$ S : U \cap Z \to X $$ where $U \subseteq X$ is open.
Question: Can I define the Leray-Schauder degree of $S$? If I had $S : U \to X$ this would be fine by the usual definition, assuming all the required conditions (compactness, no fixed points on the boundary etc) are met.
Motivation
I'm interested in the problem $$ u''(s) = f_t(u(s), s) $$ for $s \in \mathbb{S}^1$ with $f_t = f(\cdot, \cdot, t)$ and $f : \mathbb{R}^2 \times [0, 1] \to \mathbb{R}$ continuous. Assuming I can solve the problem for $f_0$, the goal is to use a degree theoretic argument to show I can solve the problem for $f_1$. The issue I am having is in defining the appropriate operators with well defined degrees.
My setting is on the Banach space, $X = C^2(\mathbb{S}^1 \to \mathbb{R})$ equipped with the usual $C^2$ norm. Let $j : C^2(\mathbb{S}^1 \to \mathbb{R}) \to C^0(\mathbb{S}^2 \to \mathbb{R})$ be the (compact) inclusion where $C^0$ is equipped with the sup-norm. Letting $L(u) = u''$, $F_t : C^0(\mathbb{S}^1 \to \mathbb{R})$ be given by $F_t(u)(s) = f(u(s), s)$, the problem becomes $$ L(u) = F_t \circ j(u) $$ which I rewrite as a fixed point problem $$ u = S_t(u) $$ with $S_t = L^{-1} \circ F_t \circ j$ which will be a compact map provided I work on a suitable domain where $L$ is invertible. For the sake of argument, we can assume that I'm working in a bounded open set $U \subseteq X$. Then the Leray-Schauder degree $\operatorname{deg}(I - S_t, U, 0)$ will be defined provided there are no fixed points on the boundary and then a solution exists if and only if $\operatorname{deg}(I - S_t, U, 0) \ne 0$.
The issue I have is with $L^{-1}$. The kernel of $L$ is the constant maps and it has a closed complement $$ \tilde{X} = \left\{u \in X : \int u ds = 0\right\} $$ Of course $\tilde{X}$ is precisely the range of $L$, so $L : \tilde{X} \to \tilde{X}$ is invertible self map of the Banach space $\tilde{X}$. Great!
But for $u \in \tilde{X}$, there is no reason why $F (u) \in \tilde{X}$ should be true, hence $L^{-1} \circ F$ is not defined. I could restrict to the closed set $$ Z = F^{\ast} \tilde{X} \cap \tilde{X} = \{u : F(u) \in \tilde{X}\} \cap \tilde{X}. $$ Then $L^{-1} \circ F$ is defined, but now $$ S_t : Z \cap U \to \tilde{X}. $$ which leads to my question.
I can answer my own question after all:
Define $M_t$ mapping $C^2(\mathbb{S}^1 \to \mathbb{R}^2)$ to itself by $$ M_t = P + [P + L^{-1} \circ (I - P)] \circ F_t $$ where $$ P(X) = \frac{1}{2\pi} \int_{\mathbb{S}^1} X ds $$ is the projection onto $\ker L$ via the splitting $C^2(\mathbb{S}^1 \to \mathbb{R}^2) \simeq \ker L \oplus \operatorname{R}(L)$. Then $I - P$ is the projection onto $\operatorname{R}(L)$. Thus $L^{-1} \circ (I - P) \circ G$ is well defined and there's no need to restrict to any subspaces like $Z$ in the original question.
Then it's straight forward to show that $$ L(X) = F_t(X) $$ if and only if $$ X = M_t(X) $$ and hence I just need to study the degree of $M_t$.
This construction may be found in:
"The Leray-Schauder degree of $S^1$-equivariant operators associated to autonomous neutral equations in spaces of periodic functions" Journal of Differential Equations, Volume 92, Issue 1, 1991, Pages 90-99, ISSN 0022-0396, https://doi.org/10.1016/0022-0396(91)90065-H