$C[0,1]/C_0$ is isomorphic to $\Bbb R$

Question

Let $$ C_0= \{f \in C[0,1] \mid f(1/2)=0 \} $$

then $ C_0$ is a subspace of $C[0,1]$ and $C[0,1]/C_0$ is isomorphic to $\Bbb R$. I have proved that $ C_0 $ is a vector subspace. I tried to prove that $C[0,1]/C_0$ is isomorphic to $\Bbb R$.

I know that

$ C[0,1]/C_0 = \{ f+C_0 : f \in C[0,1] \} $

I tried this function:

$ h: C[0,1]/C_0 \to\Bbb R $

$ h( f+C_0)=\max \{f(x) :x \in [0,1]\} $

but this is not a bijection. So how can I prove that?

Answer 1

For such a question (prove that $F$ is a subthing of $E$ and that $E/F$ and $G$ are isomorphic as things, where "thing" may be replaced for instance with "group" or "vector space"), use systematically the first isomorphism theorem. In your case:

If $\varphi:M\to N$ is a homomorphism of vector spaces (i.e. a linear map), then $\ker\varphi$ is a subspace of $M$, $\operatorname{im}\varphi$ is a subspace of $N$, and the map $M/\ker\varphi\to\operatorname{im}\varphi,\;m+\ker\varphi\mapsto\varphi(m)$ (is well defined and) is an isomorphism.

(If you weren't taught this theorem yet, I warmly encourage you to prove it, as an educational exercise and a promising investment.)

Applying this theorem to the two vector spaces $M=C[0,1]$, $N=\Bbb R$ and to the linear map $$\varphi:C[0,1]\to\Bbb R,\;f\mapsto f(1/2)$$ instantly proves that $C_0$ is a subspace of $C[0,1]$, since it is the kernel of $\varphi$. Moreover, $\varphi$ is onto ($\varphi(f_r)=r$ where for any $r\in\Bbb R$, $f_r$ denotes the constant function $x\mapsto r$). Therefore, the map $$C[0,1]/C_0\to\Bbb R,\;f+C_0\mapsto f(1/2)$$ is an isomorphism.

Remark 1. Your map $$ h: C[0,1]/C_0 \to\Bbb R,\;f+C_0\mapsto\max \{f(x) :x \in [0,1]\}$$ was not well-defined, because the max may be different for two functions in the same class. For instance, if $f(x)=0$ and $g(x)=x-\frac12$ then $f,g\in C_0$ hence $f+C_0=C_0=g+C_0$ but $\max f([0,1])=0\ne\frac12=\max g([0,1])$. Besides, even if you don't quotient by $C_0$, i.e. if you define $$k: C[0,1]\to\Bbb R,\;f\mapsto\max \{f(x) :x \in [0,1]\},$$ you will have the problem that $k$ is not linear.

Remark 2. I followed your wish to stay in the context of vector spaces, but you should be aware that there is an analogous isomorphism theorem for rings, which proves (since our $\varphi$ is not only linear but satisfies $\varphi(fg)=\varphi(f)\varphi(g)$, and $\varphi(f_1)=1$ as a special case of $\varphi(f_r)=r$) that $C_0$ is an ideal of $C[0,1]$ and our map $C[0,1]/C_0\to\Bbb R$ is also an isomorphism of rings (hence finally: of $\Bbb R$-algebras).