Let $X$, $Y_1$, $Y_2$ be Banach spaces (if that helps I will be happy to assume they are Hilbert spaces) and let
\begin{align}
T_1 : X \rightarrow Y_1, \\
T_2 : X \rightarrow Y_2
\end{align} be bounded linear maps such that
$T:= T_1 \times T_2 : X \rightarrow Y_1 \times Y_2$ is a Fredholm map.
Is it true that $T_1|_{\ker T_2}$ is also Fredholm? Its kernel is finite-dimensional since it is the kernel of $T$ but why is the cokernel finite-dimensional?
Best Answer
Let $\mathcal{P}(n)$ be the property: "For all Banach spaces $X$, $Y_1$ and $Y_2$, and for all $T_1 \in B(X, Y_1)$ and $T_2 \in B(X, Y_2)$, if $T_1 \times T_2$ is Fredholm with $\mathrm{codim}(\mathrm{ran}(T_1 \times T_2)) = n$, then $T_1|_{\ker T_2}$ is Fredholm."
As noted by OP, we need only focus on the cokernel or on the range (just a matter of preference) as the kernel of $T_1|_{\ker T_2}$ is already finite-dimensional.
Initialisation with $n = 0$: fairly simple, $\mathrm{codim}(\mathrm{ran}(T_1 \times T_2)) = 0$ is equivalent to saying that $T_1 \times T_2$ is surjective, thus its range contains $Y_1 \times \{0\}$, but that implies that $T_1|_{\ker T_2}$ is also surjective, thus Fredholm by the previous remark.
Induction phase: assume $\mathcal{P}(n)$ is true, and let's show that $\mathcal{P}(n+1)$ is also true.
Let $X$, $Y_1$, $Y_2$, $T_1$, and $T_2$ be as they should, with $\mathrm{codim}(\mathrm{ran}(T_1 \times T_2)) = n + 1$.
Define $\widetilde{X} := X \times \mathbb{K}$ ($\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$), equipped with one of the usual product norms. Take $V$ an algebraic complement of $\mathrm{ran}(T_1 \times T_2)$ in $Y_1 \times Y_2$, let $(y_1,y_2) \in V$, and consider $V_0$ an algebraic complement of $\mathrm{span}\big((y_1, y_2)\big)$ in $V$, so that: $$Y_1 \times Y_2 = \mathrm{ran}(T_1 \times T_2) \oplus \mathrm{span}\big((y_1, y_2)\big) \oplus V_0$$ with $V_0$ of dimension $n$, as $V$ is of dimension $n+1$ by assumption.
Consider now the linear maps $S_1 : \widetilde{X} \to Y_1$ and $S_2 : \widetilde{X} \to Y_2$ defined as follows: $$\forall (x, \alpha) \in \widetilde{X},\quad (S_1 \times S_2)(x, \alpha) := (T_1 \times T_2)(x) + \alpha(y_1, y_2)$$ $S_1$ and $S_2$ are linear, and bounded because: $$\|S_k(x,\alpha)\| = \|T_k(x) + \alpha y_k\| \leq \|T_k\|\|x\| + |\alpha| \|y_k\| \leq \max(\|T_k\|, \|y_k\|) \|(x,\alpha)\|$$ Moreover, $\ker(S_k) = \ker(T_k) \times \{0\}$, thus: $$\ker(S_1 \times S_2) = \left(\ker(T_1)\times\{0\}\right) \cap \left(\ker(T_1)\times\{0\}\right) = \ker(T_1 \times T_2)\times\{0\}$$ and: $$\mathrm{ran}(S_1 \times S_2) = \mathrm{ran}(T_1 \times T_2) \oplus \mathrm{span}\big((y_1, y_2)\big)$$ Hence, $S_1 \times S_2$ is Fredholm because $T_1 \times T_2$ is Fredholm, and $\mathrm{ran}(S_1 \times S_2)$ is complemented by $V_0$ of dimension $n$, therefore, by $\mathcal{P}(n)$, $S_1|_{\ker S_2}$ is Fredholm.
Now, we have: $$\mathrm{ran}\left(S_1|_{\ker S_2}\right) = \mathrm{ran}\left(S_1|_{\ker T_2 \times \{0\}}\right) = \mathrm{ran}\left(T_1|_{\ker T_2}\right)$$ By the observation made prior to the induction, this implies that $T_1|_{\ker T_2}$ is Fredholm because $S_1|_{\ker S_2}$ is Fredholm, which means that $\mathcal{P}(n+1)$ holds assuming $\mathcal{P}(n)$ holds.
Conclusion: By induction, $\mathcal{P}(n)$ is true for all $n \in \mathbb{N}$, hence OP's claim holds.