Why T-duality only work when the background has isometries? I have been studying from some textbooks and papers about the T-dality topic. In particular for the Buscher rules it seems that they claim that in order to have T-duality in certain direction we need to have an isometry in that direction. For example in [1] they say:

we can now give a more
systematic discussion of T-duality, which is valid whenever the background has isometries.

and in [2]:

One requires the metric to admit at least one continuous abelian isometry leaving invariant the $\sigma$-model action constructed out of $(g, b, \phi)$.

Also I read the Buscher's paper [3] and he says:

If this manifold admits the
action of an holomorphic isometry, the model is dual...

So this makes me think:
Are isometries necesary for T-duality? or Is T-duality necesary for isometries(in the sense that if we have an isometry it is guaranteed to have a T-duality)?

[1] Blumenhagen, Lust, Theisen; Basic Concepts of String Theory; page 432
[2] Alvarez, Alvarez-Gaume, Lozano; An Introduction to T-Duality in String Theory; Nucl.Phys.Proc.Suppl.41:1-20,1995 arXiv:hep-th/9410237  page 3
[3] BUSCHER; A SYMMETRY OF THE STRING BACKGROUND FIELD EQUATIONS; Phys. Lett. B 194 (1987) 59
 A: For abelian T-duality, the answer is yes, you need isometries. Gauging these isometries is how you construct your T-dual, so without isometries I don't see how you would construct the T-dual background.
For non-abelian duality, the answer is...not really, but it's a bit more complicated.
Non-abelian T-duality generalises the usual T-duality to the action of a (generically) non-abelian group G. This group acts on your background by isometries, and gauging this isometry gives you a non-abelian T-dual background. Unlike the case of abelian T-duality, the dual background will generically have fewer isometries than your original background (and may indeed have no isometries), and in particular, performing the same procedure of gauging isometries to obtain your original background isn't possible (except possibly is exceptional cases). For this reason, and because it's not expected that non-abelian T-duality is a full symmetry of string theory, it's somewhat of a misnomer to call it a "duality" (but we're stuck with that name for historical reasons).
So with non-abelian T-duality, you have two spaces related by a duality. Starting from your original background, you can construct the dual background easily enough, but if you were to start with the dual background, there would be no way of constructing your original background because there is no isometry group to gauge. This seems to suggest that there is an overarching framework governing T-duality, which sometimes manifests itself as isometry groups. There is, and it's called Poisson Lie T-duality.
Poisson Lie T-duality generalises non-abelian and abelian T-duality further. It starts with an algebraic structure called a Lie bialgebra (which is just a Lie algebra whose dual space also has a Lie algebra structure). The two algebras of this bialgebra, (g,g'), determine the nature of the duality, and are determined by the sigma model (and it's dual). T-duality then corresponds to interchanging the roles of g and g'. 
When the Lie algebra structure, g', on the dual space is abelian, then g is the Lie algebra of a group of isometries acting on your original background. So if both g and g' are abelian, this corresponds to the usual abelian T-duality. If g is non-abelian and g' is abelian, then your original background has a non-abelian group of isometries acting on it, but the dual background does not (since we now have (g',g) and the Lie algebra structure on the dual space is non-abelian, so we can't think of g' as the Lie algebra for a group of isometries). 
If both g and g' are non-abelian, then neither group acts as a group of isometries on either the original or dual background, yet they are nevertheless related as sigma models. In this sense, you can have T-duality without isometries.
