Problem: Let's consider a linear space of smooth functions $C^k[a,b]$ and a submersion $g:C^k[a,b]\to\mathbb R$. $M = \{x\in C^k[a,b] | g(x) = 0\}$ is a differentiable manifold. What is the tangent space of $M$ at any given point?
My thoughts: As in a finite-dimensional case my hope is that the answer would be $T_aM = ker g'(a)$ (see tangent space of manifold and Kernel for example). I think the proof should look something like this:
Let's take any vector $y$ from $T_aM$ and any of its tangent curves $\gamma:\mathbb R \to M$ with $\gamma(0) = a$. As for any $x\in M$, $g(x)=0$ it follows that $g \circ \gamma \equiv 0$, so $(g \circ \gamma)' = 0$ and $g'(y) = 0$. So inclusion $T_aM \subset ker g'(a)$ is proven.
As $T_aC^k[a,b]\over ker g'(a)$ $\simeq \mathbb R $ by fundamental homomorphism theorem, we know the dimension of $T_aC^k[a,b]\over ker g'(a)$. Now it's for the trickiest part: we have to somehow find the dimension of $T_aC^k[a,b]\over T_aM$ and prove that as dimensions of those factors are equal and there is an inclusion from (1), the linear spaces in denominators are the same. For now I have no idea how to do that.
Any help is appreciated.