Zariski topology and morphisms that coincide on a dense set

Question

In Miles Reid Undergraduate Algebraic Geometry book it is stated informally, about the Zariski topology:

(1) two morphisms which coincide on a dense open set coincide everywhere

I am suprised that he requires a dense open set since morphisms are continuous maps for the Zariski topology and I thought that for a general topological space we had the following statement

(2) two continuous maps which coincide on a dense set coincide everywhere

My question is: is (2) a valid statement of general topology? If not, what specificities of the Zariski topology make it invalid? (compared to, for instance, a metric space — which I'm more familiar with)

EDIT: Also, a link to a proof of the correct statement, with the precise conditions, would be much appreciated

Answer 1

Even the first statement is false in general, consider the map from $\mathbb{A}^1_k$ to the line with double origin, you can map the origin from your source to either of the origins of the target and get two different maps that agree on $\mathbb{A}^1_k \setminus \{0 \}$. However the statement holds as long as your source is reduced and your target is separated.

This already tells you what you want to assume for a similar result in point-set topology. Recall that separateness resembles the Hausdorff property and it is in fact true that if two continous maps $f: X \to Y$ between any topological space with Y Hausdorff agree on a dense set, then they agree everywhere

Answer 2

Actually, I don't think either of these two claims are true, at least in a general-enough setting.

Consider a topological space $X$ obtained by gluing two copies of $\mathbb{R}$ along $\mathbb{R}\setminus\{0\}.$ It looks like a line with a double origin. It comes with two natural embeddings of $\mathbb{R}$ different in the choice of 'origin' to map the point 0 to. Nevertheless, these two maps agree on a dense set. This construction can be emulated in algebraic geometry once you define a sufficiently general type of spaces, such as schemes.

The statement becomes true once you require the target space of your pair of morphisms to be Hausdorff. In the world of algebraic geometry this translates to requiring the source to be reduced and the target to be separated. I suspect that Reid only covers projective and affine varieties over an algebraically closed field, where this is automatic.

Answer 3

In the statement $(1)$, the word morphism is absolutely crucial, and the statement is false if you replace morphism by continuous for the Zariski topology. (I presume it is a a statement about varieties over an algebraically closed field, otherwise one needs additional assumptions).

The statement about general topology you have in mind is the following: if $f,g:X\to Y$ are continuous maps between topological spaces agreeing on some dense subset, and $Y$ is Hausdorff, then $f=g$. See e.g here.

Now the Zariski topology is most of the times not Hausdorff (for finite type schemes over a field this basically only holds if your space is finite), it is way to coarse for such a statement to work for any continuous maps. For example, if you take $X=Y=\mathbb{A}^1_{\mathbb{C}}$ to be the complex affine line, the non-empty open sets are precisely the sets with finite complement (i.e. the Zariski topology is the co-finite topology in this case; if we ignore the generic point). In particular, any map $f:X\to Y$ such that $f^{-1}(y)$ is finite for all $y\in Y$ is continuous. So if you take for example $f$ to be the identity, and $g$ to be the map which is the identity on $\mathbb{A}^1_{\mathbb{C}}\setminus\{0,1\}$ but such that $g(0)=1$ and $g(1)=0$, then $f,g$ are continuous and agree on a dense open set, but $f\neq g$.

So the particularity of the Zariski topology which makes this false for arbitrary continuous maps, is that it is in general way to coarse, and many maps are continuous.