In Voisin's book *Hodge theory and complex algebraic geometry, I* Section 9.1.2, p.223, the author writes:



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> Let $\phi:\mathcal X\to B$ be a family fo complex manifolds. The differential $\phi\_\*$ is a morphism of holomorphic vector bundles $T\_{\mathcal X}\to \phi^\*T\_B$.
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If my understanding is right, $\phi^\*T\_B$ is the space of pullbacks of vector fields in $T\_B$. But is it always possible to pull back a vector field? As far as I know, differential forms can always be pulled back, but I never see anyone push forward a differential form. Similarly, we can often see push forward a vector at a point, but I'm not sure pull back a vector field is always possible. In Loring. Tu's book *an introduction to manifolds, 2nd edition*, p.196, the author writes:



> 
> Unlike vector fields, which in general cannot be pushed forward under a smooth map, every covector field can be pulled back by a smooth map.
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He even did not mention the pullback of a vector field, which, I guess, implies that pull back a vector field is generally impossible, then why did Voisin write $\phi^\*T\_B$ without hesitation? Or is the submersivity of $\phi$ guarantees the existence of $\phi^\*T\_B$ tacitly?


In summary, my question is:


Under which condition, can we pull back a vector field?



Pull back a vector field
You are **not** pulling back vector fields.


You are pulling back the vector bundle $T\_B$ to be a bundle over $\mathcal{X}$. (See, e.g. <https://en.wikipedia.org/wiki/Pullback_bundle> for a description.)


Notice that in general, the nomenclature "pushforward of a vector field" is imprecise. When the mapping is not bijective, "$\phi\_\*v$" does not define a vector field over the codomain: if $q$ is not in the image of $\phi$ then $\phi\_\*v$ is undefined, and if $\phi$ is not injective the same $q$ may be associated to two distinct values of $\phi\_\*v$.


Thinking in terms of the pullback tangent bundle solves both of these issues.


(The pullback of a differential form, on the other hand, is always precise, since functions must be well defined on its domain.)


