Suppose $A$ and $B$ are two algebraic varieties, and $f:A\to B$ is a morphism of algebraic varieties. I guess it is true that $\text{im}(f)$ is itself an algebraic variety. But how to prove it?
[Math] Image of a morphism of varieties
abstract-algebraalgebraic-geometry
Related Solutions
Mariano Suarez-Alvarez's point about understanding the intuition as you learn the theory more is correct, but I'd like to give a partial answer to help guide your intuition. After all, it is possible to spend months or years learning algebraic geometry and come away with little intuition of what the whole subject is about.
First, algebraic varieties are geometric spaces which look locally like affine varieties. In this sense, the theory is developed similar to, say, the theory of manifolds where a manifold is defined to be a space that is locally Euclidean. Of course, that limits the local study of manifolds - any two manifolds are locally isomorphic. Not so for algebraic varieties, as there is a wide variety of affine varieties.
So I think you should begun by restricting your question to affine varieties. And the key is that affine varieties are completely determined by their ring of globally regular functions. In other words, two (irreducible) closed subsets of affine space are isomorphic iff we can find a global 'change of variables' that identifies the global regular functions on the two spaces. Rescaling $(x,y) \mapsto (\sqrt{2}x,\sqrt{2}y)$ yields the isomorphism between $x^2+y^2=1$ and $x^2+y^2=2$.
I'll modify your non-example (because $\mathbb{A}^2 \setminus \{0\}$ is not affine) and explain why $\mathbb{A}^1$ and $\mathbb{A}^1 \setminus \{0\}$ are not isomorphic. Their rings of regular functions are $k[T]$ and $k[T,T^{-1}]$ respectively, which are not isomorphic. So there can be no 'changes of variables' that identifies the two spaces.
One important caveat: when I say there is global 'change of variables' from $X \subset \mathbb{A}^n$ and $X' \subset \mathbb{A}^{n'}$, I am talking about using polynomial maps that are restricted from the respective affine spaces, but they only needed to be defined on the spaces $X$ and $X'$. For example $\mathbb{A}^1 \setminus \{0\}$ (viewed as $t \neq 0$) and $xy=1$ are isomorphic via $t \mapsto (t, 1/t)$ and $(x,y) \mapsto x$. Of course, $1/t$ is only a valid change of variables when $t \neq 0$, but fortunately we are only looking at points where $t \neq 0$.
The global story is similar, except that we cannot just compare globally regular functions. (For example, the only globally regular functions on any projective variety are the constant functions, yet intuitvely there ought to be many different projective varieties up to isomorphism.) So now we require a global 'change of variables' so that regular functions on local pieces match up with the regular functions on the corresponding local pieces.
I am not sure if this explanation is what you are looking for. Algebraic geometry is very much a function oriented theory. We compare spaces by looking at the functions on them. One can take such an approach to manifolds as well. But for manifolds we also have an intuition for what the possible change of variables are ('stretching' and 'twisting' and the like). It's much harder to tell such a story in algebraic geometry because algebraic varieties are so much more diverse. There are still some basic intutions such as you can't have an isomorphism between a smooth variety and a singular variety because isomorphisms give rise to (vector space) isomorphisms of tangent spaces. But there are lots of possible singularities, and getting a hold on them is a major on-going project in the field. For example, you could study plane curves in depth and learn to tell apart singularities in this case (using blowups). But then you'll quickly discover the singularities on surfaces are more complicated and those on higher dimensional varieties still more complicated and hard to get a handle on.
I will assume varieties are irreducible. Let $I$ be the kernel of $f^\sharp : \mathbb{C}[W] \to \mathbb{C}[V]$. Since $f^\sharp$ is surjective, $I$ must be a prime ideal, so corresponds to some closed subvariety $Y \subseteq W$. Thus the morphism $f : V \to W$ must factor through the inclusion $Y \hookrightarrow W$. We may assume without loss of generality that $Y = W$. But then $f^\sharp$ is an isomorphism, say with two-sided inverse $g^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$, and the fundamental theorem regarding morphisms of varieties implies the morphism $g : W \to V$ corresponding to $g^\sharp$ must be a two-sided inverse for $f : V \to W$.
You are right that the image of a morphism of affine varieties need not be closed: for example, if $V = \{ (x, y) \in \mathbb{C}^2 : x y = 1 \}$, $W = \mathbb{C}$, and $f(x, y) = x$, then the image of $f$ is a not closed in $W$. But what about $f^\sharp$? Well, $\mathbb{C}[V] = \mathbb{C}[x, y] / (x y - 1)$ and $\mathbb{C}[W] = \mathbb{C}[x]$, so $f^\sharp : \mathbb{C}[V] \to \mathbb{C}[W]$ is not surjective. (In fact, it is injective!) Thus the argument in the previous paragraph does not apply.
Best Answer
You can't prove it because it is not true!
Consider the (dominant) morphism $f\colon \mathbb C^2\to \mathbb C^2\colon (x,y)\mapsto (x,xy)$.
Its image is the subset $Im(f)=\{(u,v)\in \mathbb C^2 | u\neq 0 \} \cup \{(0,0)\}$.
This set is not locally closed in $\mathbb C^2$ and so $Im(f)$ is not a subvariety of $\mathbb C^2$.
Feel free to soup up the example by introducing arbitrary fields, schemes,... :-)