My question is this: what is the topological analog of the Grothendieck group of coherent sheaves $G(X)$?

Background:

In Algebra/Algebraic Geometry there are two versions of the Grothendieck group of a variety $X$, one is the Grothendieck group of vector bundles, which we can denote as $K(X)$ and the other is the Grothendieck group of coherent sheaves, which we can denote by $G(X)$.

As the category of vector bundles is a subcategory of coherent sheaves there is always a group homomorphism $K(X) \to G(X)$ and if $X$ is smooth then this map is an isomorphism (basically because every coherent sheaf will be resolved by a finite complex of vector bundles as soon as $X$ is regular by Serre's Theorem).

Now let us assume that $X$ is a complex variety, so that we can compare algebraic K-theory with the topological one. That is there is a ring homomorphism $K(X) \to K^{top}(X)$, typically non-injective and non-surjective.

I am asking for groups $G^{top}(X)$ which are covariantly functorial for proper morphisms, admit natural transformations $G(X) \to G^{top}(X)$ as well as natural maps $K^{top}(X) \to G^{top}(X)$, such that the latter maps are isomorphisms in the smooth case.

Some remarks:

The question makes sense for higher G-theory as well but I only bring up Grothendieck group to keep things simple and to avoid the clash of notation (e.g. $K_0(X)$ in algebra vs $K^0(X)$ in topology).

A natural guess is that what I call $G^{top}(X)$ is simply what is called K-homology, could this be the case?

Finally, there is a modern procedure of taking topological K-theory of a triangulated category due to Blanc (I think), and as far as I understand, $K^{top}(Perf(X)) = K^{top}(X)$, no matter whether $X$ is smooth or not, so it may be that what I want to take is $K^{top}(D^b(Coh(X))$, is this something sensible?