Morphism in category theory Morphisms in category theory map from one object to another object. E.g. for group, object is set, right? Homomorphism is morphism. For topplogy space, object is set. Homeomorphism is morphism.
My question is: does Morphism only map the elements in the object? Because normally, we don't only have set, but also some structures, like operators, topology. Are these structures mapped by morphisms as well?
 A: In your examples, a morphism is specified by its behavior on elements, but has to satisfy the further property of preserving some algebraic or topological structure. You wouldn't say that the group operations are mapped to each other in a group homomorphism, or if you did you'd just be restated the usual homomorphism condition. 
On the other hand, in many categories objects don't have elements at all, such as the example in the comments. So while it's extremely common to have as morphisms structure-preserving functions like in your examples, it's not required and you shouldn't bake it into your intuition about a category.
A: What you are trying to describe is a "concrete category", i.e. a category $\mathbf C$ with a faithful functor to the category of sets. I will try to introduce you to that notion. Examples are 


*

*$\mathbf{Grp}$ : Objects are groups, morphisms are group homomorphisms. A group homomorphism $\varphi : G \to H$ is characterized by its underlying set map. 

*$\mathbf{Vect}_K$ : Objects are vector spaces over the field $K$ ($\mathbb R$ or $\mathbb C$ if you like), and morphisms are linear maps. A linear map $f : V \to W$ is determined by its underlying set map. 

*$\mathbf{Top}$ : Objects are topological spaces and morphisms are continuous maps (not homeomorphisms : although "topological spaces with homeomorphisms as morphisms" would be a different category (a subcategory of $\mathbf{Top}$), it is not the one we typically use!). Again, a continuous map $f : X \to Y$ is determined by its underlying set map. 


Of course, you can also do other examples, like pointed topological spaces, rings, $A$-algebras, associative algebras, monoids, magmas, semigroups, and the list goes on and on. 
The part where I mentioned that the morphism was "determined by its underlying set map" means that if you let $U : \mathbf C \to \mathbf{Set}$ be the functor which sends the object (group, vector space, topological space) to its underlying set and the morphism to its underlying function, then if the two functions agree, the two morphisms are equal. 
Sometimes there is a natural choice of functor to $\mathbf{Set}$ but it is not faithful ; a very famous example (if you know some algebraic geometry) is the category of irreducible separated schemes of finite type over an algebraically closed field $k$ (say $\mathbb C$ if you want), with morphisms being the $\mathrm{Spec}(k)$-morphisms of schemes.  In this case, a functor to $\mathbf{Set}$ is given by first sending $f : X \to Y$ to $f_{\mathrm{red}} : X_{\mathrm{red}} \to Y_{\mathrm{red}}$, which means that $X_{\mathrm{red}}$ and $Y_{\mathrm{red}}$ are varieties, and there is a faithful functor $\mathbf{Var} \to \mathbf{Set}$. This is the "obvious choice" of functor to $\mathbf{Set}$, but it is not faithful because reducing the morphism removes the information given by "admitting nilpotents" (when we reduce, in the case of an affine scheme, we essentially mod out the nilradical). 
Most categories you encounter at the beginning are concrete and you should work out what the functor $U$ should be, it is a good exercise. However, after a while you start encountering categories which are not concrete, i.e. in which you are not given a faithful functor to $\mathbf{Set}$, let alone a functor at all. In other words, you stop thinking as your objects as some sets with extra structure because that's not what they are. You start enjoying this generality when you encounter a few more examples though!
EDIT : Category theory does not answer that question. What category theory does is consider the collection of all your objects and morphisms that you already defined and consider them as a whole, remembering only what morphism is the identity and how to compose morphisms. That doesn't even assume that the objects and the morphism admit a notion of "structure" that could be "mapped". We need to define that part outside of category theory.
What you are trying to say is if assuming we have a morphism $f : M \to N$, does the morphism $f$ send some structure in $M$ to the structure in $M$. That is not the correct way to think about it. The correct way is to think that morphisms respect the structure of both the domain and codomain. In the case of algebraic structures this often presents in the form
$$
f(m \star n) = f(m) \star f(n).
$$
In the case of manifolds, the map preserves the local structure, that is, for $f : M \to N$, at every point $x \in M$, there exists charts $U,V$ around $x$ and $f(x)$ where $f$ "looks like a smooth map between $\mathbb R^m$ and $\mathbb R^n$. It does not send the structure (even the local one) of $M$ to that of $N$ ; that would require the morphism to be an injective immersion, in which case $f(M)$ would correspond to a submanifold.
Hope that helps,
