Kernel of a function? The kernel of a linear map (=matrix) is all vectors that map to $0$. 
If I have a function $f:X \to Y$ where $X,Y$ are topological spaces, when does it hold that $\ker f = \{0\} \iff f$ is injective? Assume that the kernel of $f$ is all $x$ such that $f(x) = 0$.
It clearly does not hold for arbitrary $f$ because if $X = Y = \mathbb R$ and $f(1) = 2$ and $f(x) = x$ for $x \neq 2$ then the kernel is trivial but $f$ is not injective. But what about continuous $f$?
 A: 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 $\ker f = \{0\}$.
A: For one thing, "$0$" is not assumed to be an element of a general topological space.  In a vector space, $0$ is the name for the additive identity.  So in most cases, the question is not well formed.
If $X$ and $Y$ are topological vector spaces (or abelian groups), then there are designated "$0$"s in $X$ and $Y$.  But:


*

*General continuous maps between such spaces need not map $0$ in $X$ to $0$ in $Y$.

*A continuous map that sends $0$ in $X$ to $0$ in $Y$, and sends no other element of $X$ to $0$ in $Y$, need not be injective.
Examples:  


*

*$X=Y=[1,2]$.  No zeros in sight, regardless of $f$.

*$X=Y =\mathbb R$, $f(x)=x+1$ is injective but does not send $0$ to $0$.  It does send $-1$ to $0$.

*$X=Y=\mathbb R$, $f(x) = x^2+1$ is not injective, does not send $0$ to $0$, and for that matter doesn't send anything to $0$.

*$X=Y=\mathbb R$, $f(x)=e^x$ is injective, and also doesn't send anything to $0$.

*$X=Y=\mathbb R$, $f(x)=x^2$ is not injective, but $\{x\in X:f(x)=0\}=\{0\}$.


In general, $f$ is injective if for all $y\in Y$, the set $f^{-1}\{y\}=\{x\in X:f(x)=y\}$ has at most one element.  This reduces to the special case $f^{-1}\{0\}=\{0\}$ in the case where $X$ and $Y$ are vector spaces (or abelian groups) and $f$ is a linear transformation (or group homomorphism).  Since $0$'s importance is relative to algebraic structure, you should expect that $f$ must preserve some algebraic structure (e.g., by being linear) in order to get a criterion using $0$.
