$X = C[0,1]$ and $M = \{f | f(0) = 0\}$, then $X/M \cong \mathbb{C}$.

Question

Let $X = C[0,1]$ and $M = \{f | f(0) = 0\}$. I would like to show that $X/M \cong \mathbb{C}$. This is a problem in Reed & Simon's book on functional analysis.

How would one start such a problem? I am thinking of using the fact that $X$ and $M$ are vector spaces so that I can use the first isomorphism theorem. I would need to construct a homomorphism $\phi: C[0,1] \rightarrow \mathbb{C}$ such that Ker$(\phi) = M$.

Define $\phi(f) = f(0)$. Then $$\phi(fg) = (fg)(0) = f(0)g(0) = \phi(f)\phi(g)$$ so $\phi$ is a homomorphism. Since constant functions are continuous, $\phi$ is a surjection.

Is this the right approach or is there another way?

Answer 1

Consider: $\Phi: C[0, 1]/M\to\Bbb{C}$ defined by $$\Phi(\hat{f})=f(0) $$

Claim:

1. $\Phi$ is well defined.

2. $\Phi\in (C[0, 1]/M)^{\text{#}}$ i.e $\phi$ is a linear functional on $C[0, 1]/M$

3. $\phi$ is one-to-one and onto.

Hence $\Phi$ is a linear Isomorphism /vector space homomorphism.