A topological space $X$ is called k-space if the following condition holds:
$A\subseteq X$ is open in $X\iff A\cap K$ is open in $K$ for any compact subest $K$ of $X$.
A space $kX$ is a topological space $X$ equipped with k-topology, denoted by $\mathcal{O}(kX)$, i.e.,$U\in\mathcal{O}(kX) \iff U\cap K$ is open in $K$ for any compact subest $K$ of $X$.
Let $f:X\rightarrow Y$ be a continuous mapping from k-space $X$ to a topological space $Y$. Can we induce a continuous mapping $g:kX\rightarrow Y$ from a topological space $kX$ to $Y$?
I think that $g=f$ but I do not know whether $g$ is continuous mapping. Can someone help me?
Your question is worded in a way that makes the answer trivial, so I will assume that you mean to ask:
"If $f: X \to Y$ is continuous, then is $f: kX \to Y$ necessarily continuous?"
This is indeed true, since the $k$-topology is always finer than the original topology. This is mentioned in the nlab article, for instance, and probably also in Brown's topology book.
If you have a continuous map $f : X \to Y$ and you pick a finer topology on the domain, then the continuity of the map is unchanged. It is only if you make the topology on the codomain finer that you might cause the map to become discontinuous since there will be more open sets whose preimages will have to be open.