Without assuming some algebraic structure on $X$ and $Y$, your question is fairly meaningless, because even if $X$ and $Y$ do contain points called $0$, those points are not topologically special in any way.
As Chris Eagle has hinted, the definition of kernel that you are using is not appropriate for general functions. In the general situation, we define the kernel of $f:X\rightarrow Y$ to be the equivalence relation on $X$ given by $x\sim y$ iff $f(x) = f(y)$.
If $f$ is a linear transformation (or more generally, a homomorphism), then the kernel can be recovered just by looking at the equivalence class of the identity in $X$ under the kernel relation, and so it's simpler (especially for undergraduates who may not have learnt about equivalence relations yet) to consider the kernel as being the equivalence class of $0$, that is $\{x\in X \mid f(x) = f(0) = 0\}$.
So, once we've defined the kernel as an equivalence relation, then we see that $f:X\rightarrow Y$ is injective iff the kernel is the diagonal relation (i.e. the relation $\sim$ such that $x\sim y$ iff $x=y$). In the case of $f$ being a linear transformation, this is equivalent to $\ker f = \{0\}$.