We define $Hom_{top-k-alg}$ ($\hat{R}$,$A$) to be the set of $k$-algebra homomorphism from $\hat{R}$ to $A$ which are continuous. $\hat{R}$ is the completion $\hat{R}$ $=$ $k[[x]]$ of $R = k[x]$ with respect to $J$ $=$ ($x$) and $k$ is a field. $A$ is a $k$-algebra and a topological ring. Suppose that $A$ is a Hausdorff. Show that $\phi$ is injective where
$\phi$ : $Hom_{top-k-alg}$ ($\hat{R}$,$A$) $\rightarrow$ $A$; $f$ $\rightarrow$ $f(x)$.
What I have done so far. Suppose that $\phi$ is not injective we are gonna end up to a contradiction about A being a Hausdorff.
My first attempt was since $\phi$ is not injective we can assume that $ker$ $\phi$ $\neq$ $\{$0$\}$ and then try to find $x_1$, $x_2$ $\in$ $A$ such that $\exists$ $U_1$ , $U_2$ open sets with $x_1$ $\in$ $U_1$ , $x_2$ $\in$ $U_2$ with $U_1$ $\cap$ $U_2$ $\neq$ $\varnothing$ .But really didn't work
My second attempt is to go with the uniqueness of the limit in a Hausdorff. Suppose that $x_1$,$x_2$ $\in$ $ker$ $\phi$ with $x_1$ $\neq$ $x_2$ therefore $\phi(x_1)$ $=$ $\phi(x_2)$ $=$ $0$. So if I can find a sequence $x_n$ such that $x_n$ $\rightarrow$ $x_1$ and $x_n$ $\rightarrow$ $x_2$ and I got my contradiction so I can assume that $ker$ $\phi$ is a singleton and what is left is to show that the unique element is zero. But I cannot think something about the existence of the sequence and I believe that I haven't completely understand the structure of $\phi$ since we are talking about homomorphisms $ker$ $\phi$ = $f$ with $f=0$.
I am convinced that I have to work with the contradiction but I do miss some thing so I cannot put my thoughts into a better structure so can you please light the things up a little bit? Thank you