$G$ group $f:[0,1]\rightarrow\mathbb{R}$ prove that $G/N\cong (\mathbb{R,+})$

Question

Let $G$ be the group of all the functions such that $f:[0,1]\rightarrow\mathbb{R}$, Let define for all $t\in[0,1]$ that $(f+g)(t)=f(t)+g(t)$, I need to prove that $G/N\cong (\mathbb{R,+})$, such taht $N=\{f\in G|f(1)=0\}$

My attempt:

$f$ is homomorphism because $(f+g)(t)=f(t)+g(t)$

$Im(f)=\mathbb{R}$

Now all I need is to find $Ker(f)$, I think that $Ker(f)$ is $\{f\in G |f(a\in[0,1])=0\}$ but why it should be $Ker(f)=N$?

Answer 1

You should use the homomorphism $\varphi: G\to \mathbb R$ given by $\varphi(f)=f(1)$. This is surjective and has as kernel $N$. By the way, this shows also that $G/N \cong \mathbb R $ as vector spaces, as $\varphi $ is linear and hence $N$ is a subvector space.

Answer 2

Let: $\varphi :G\to \mathbb R$ which: $\varphi (f)=f(1)$. $\varphi$ is homomorphism from $G$ to $(\mathbb R,+)$ and for every $r\in \mathbb R$, the constant function $f(x)=r$. Finally, $ker(\varphi)=N$.