A topological space $X$ is called compactly generated if following condition holds:
$A\subseteq X$ is open in $X$ iff for every compact $K\subseteq X$, $A\cap K$ is open in $K$.
My lecturer said that a topological space $X$ is called compactly generated if following condition holds:
$A\subseteq X$ is open in $X$ iff for every compact $K$ and continuous map $f: K\rightarrow X, f^{-1}(A)$ is open in $K$.
the necessary condition is clear. To prove the sufficient, I bit confuse. I think if $f: K\rightarrow X$ is continuous and $f^{-1}(A)$ is open in $K$, obviously that $A$ is open in $X$. Hence we no need condition $X$ to be a compactly generated.
Please tell me if it is not true.
The definitions are equivalent because, for a set $A\subseteq X$, the intersection $A\cap K$ is open in $K$ for every compact set $K\subseteq X$ if and only if $f^{-1}(A)$ is open in $L$ for every map $f$ from a compact space $L$ to $X$.
For one direction you need the fact that the image of a compact space under a continuous map is compact. For the other direction try to express the intersection $A\cap K$ as the preimage under some map $i:K\to X$.