There are various incompatible definitions of $k$-space/compactly generated space in the literature, as explained in wikipedia and many posts on this site. But let's focus on Definition 2 from the wikipedia article, which is the one commonly used in algebraic topology.
A space $X$ is called compactly generated (CG2) if its topology is determined by continuous maps from arbitrary compact Hausdorff spaces. In other words, it is a space such that a subset $A\subseteq X$ is closed if it is $k$-closed in the sense that $u^{-1}[A]$ is closed in $K$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.
Is that the same as requiring that the topology on $X$ be coherent with the family of images of compact Hausdorff spaces under continuous maps?
And at the level of sets, given a topological space $X$, are these two conditions on a subset $A\subseteq X$ equivalent?
- (1) $u^{-1}[A]$ is closed in $K$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.
- (2) $A\cap u[K]$ is closed in $u[K]$ for every compact Hausdorff space $K$ and every continuous map $u:K\to X$.
Clearly (2) implies (1). Can you have (1) without (2)?
The one point compactification of the Arens-Fort space seems to be a counter-example. It is not $CG_2$, but it is itself the image of a map from $\omega+1$, so certainly its topology is coherent with the images of maps from compact Hausdorff spaces, as these images include the space itself.
(To define such a map, simply let $x_n$ be any enumeration of the Arens-Fort space, and define $f(n)=x_n$, $f(\omega)=\infty$).
Remark.
To unwind this explicitly at the level of sets, if $Y$ is the Arens-Fort space and $X=Y^*$ its one-point compactification, we can see that the set $A=Y^*\backslash \{(0,0)\}$ is closed in the sense of (1), but not (2). To see this, note that if $u\colon K\to Y^*$ is a map from compact Hausdorff $K$, then $u^{-1}(\{(0,0)\})$ and $u^{-1}(\{\infty\})$ are disjoint closed sets, so by normality there is a neighborhood $U$ of $u^{-1}(\{(0,0)\})$ with $\overline{U}\cap u^{-1}(\{\infty\})=\emptyset$, hence $u(\overline{U})$ is a compact subset of $Y$, and therefore finite, as $Y$ is anticompact.
In particular, $u(\overline{U})$ is discrete, so $u^{-1}(\{(0,0)\})$ is open in $\overline{U}$ (either it is the pre-image of an isolated point in $u(\overline{U})$, or it is empty), hence open in $U$, hence open in $K$ (since $U$ is open), so $Y^*\backslash \{(0,0)\}$ is closed in the sense of (1). But since $Y^*$ is the image of a compact Hausdorff space, and $(0,0)$ is not isolated in $Y^*$, $Y^*\backslash \{(0,0)\}$ is not closed in the sense of (2).
(This is all really just unpacking the argument from this answer.)