Zariski topology and morphisms that coincide on a dense set 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

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.
