$I$-adic topology on a topological $k$-algebra $R$. Let $k$ be a field. Let $k[[x]]$ be the completion of $k[x]$ with respect to the ideal $(x)\subset k[x]$. Let $R$ be a topological $k$-algebra that is complete for the $I$-adic topology induced by some ideal $I\subset R$. We define $\text{Hom}(k[[x]],R)$ to be the set of all continuous $k$-algebra homomorphisms from $k[[x]]$ to $R$. We do now define the map 
$\varphi:\text{Hom}(k[[x]],R)\rightarrow R, f\mapsto f(x)$. 
I want to show that $I\subset \text{im}(\varphi)$. 
My approach was by assuming that there exists an $r\in I$ such that for all $f\in \text{Hom}(k[[x]],R)$ we have $f(x)\neq r$, and then using the fact that since $R$ is complete it is Hausdorff to find a contradiction. One can namely show that since $R$ is Hausdorff the map $\varphi$ has to be injective. So I hoped that I encountered a contradiction with the injectivity, but unfortunately I didn't. 

Let $r \in I$. Then $\sum_{k \geq 0}{a_kx^k} \in k((x)) \longmapsto \sum_{k \geq 0}{a_kr^k} \in R$ is (by completeness) a well-defined continuous morphism of $k$-algebras mapping $x$ to $r$. 
