In Example 3.3.1 Chapter II Hartshorne, he considers a morphism of schemes $F : X \to Y$, where $X = Spec \ k[x,y,t]/(ty – x^2)$ and $Y = Spec \ k[t]$. And he says $f$ is a surjective morphism. I was wondering what is the definition of surjectivity for morphisms of schemes? Thank you
Definition of a surjective morphism of schemes
algebraic-geometrydefinitionschemes
Related Solutions
1) The correct definition of "surjective" for a morphism of schemes $f:X\to Y $ is that the underlying map of sets $|f|:|X|\to |Y|$ be surjective.
"Correct" means "as decreed by Grothendieck" : Cf. EGAI, Chap. I, Prop 3.5.2 .
2) The categorical notion mentioned in your point (2) is called epimorphism.
3) If $\overline{f(X)} = Y$, we say that $f$ is dominant.
This is much weaker than surjectivity (even for reduced schemes), as witnessed by the inclusion of any dense open strict subset of a scheme, say $\mathbb A^1\setminus \{0\}\hookrightarrow \mathbb A^1$.
4) Surjective morphisms needn't be epimorphisms:
Let $k$ be a field and let $X=Spec(k)$, $Y=Spec(k[T]/(T^2))=k[\epsilon]$ be respectively the simple point and the double point over $k$.
Let $f:X\hookrightarrow Y$ be the closed immersion of the simple point into the double point corresponding to the $k$-algebra morphism $k[\epsilon]\to k$ sending $\epsilon$ to $0$.
The two $k$-algebra morphisms $k[\epsilon]\to k[\epsilon]$ sending $\epsilon$ to respectively $0$ and $\epsilon $ give rise to two morphisms $g_1, g_2:Y\to Y$ satisfying $g_1 \circ f=g_2 \circ f$ but $g_1 \neq g_2$.
Hence the scheme morphism $f:X\to Y$ is not an epimorphism although it is surjective, as are all maps between singleton sets.
5) Epimorphisms needn't be surjective (?):
Let $Y=\mathbb A^1_k$ (with $k$ an algebraically closed field) and $X$ be the discrete scheme obtained from the disjoint union (=coproduct) of all the closed points of $Y$.
The natural morphism $f:X\to Y$ has image $Y$ minus the generic point of $\mathbb A^1_k$ and is thus not surjective.
However it should be an epimorphism, but I haven't written out a proof of that.
I figured I would make an answer out the comments to the first answer addressing a point which confused me when I first learned this stuff. A morphism $f:X\to Y$ (I have to write it in this direction or else I'll confuse myself) is said to be a closed immersion if $f$ induces a homeomorphism of $X$ onto a closed subset of $Y$, and $f^\sharp:\mathscr{O}_Y\to f_*(\mathscr{O}_X)$ is surjective.
In some references I've seen it is casually remarked that the second condition is equivalent to surjectivity of the map $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for all $x\in X$. But is this really trivial? No! This map, which might reasonably be called the stalk of the morphism $f$ at $x$, is not literally the same as the stalk of $f^\sharp$ at $f(x)$. Indeed, that is a map $f_{f(x)}^\sharp:\mathscr{O}_{Y,f(x)}\to(f_*\mathscr{O}_X)_{f(x)}$. In general, there is always a natural map $\varphi_x:(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$, but it isn't in general an isomorphism. The map $f_x^\sharp$ is equal to $\varphi_x\circ f_{f(x)}^\sharp$. So while it is standard that the map $f^\sharp$ of sheaves (on $Y$!) is surjective if and only if $f_y^\sharp:\mathscr{O}_{Y,y}\to (f_*\mathscr{O}_X)_y$ is surjective for all $y\in Y$, this does not obviously say anything about surjectivity of the maps $f_x^\sharp:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ for $x\in X$. If however $f$ is a homeomorphism onto a closed subset $f(X)\subseteq Y$, then the stalks of $f_*\mathscr{O}_X$ at points of $Y$ are easy to compute: they are zero at points outside of $f(X)$, and at a point $f(x)\in f(X)$, we have that the natural map $(f_*\mathscr{O}_X)_{f(x)}\to\mathscr{O}_{X,x}$ is an isomorphism. So in that case, surjectivity of each $f_x^\sharp$, $x\in X$, actually will imply surjectivity of $f_y^\sharp$, $y\in Y$, and hence of $f^\sharp$.
Without the condition that $f$ is a closed topological immersion on the underlying topological spaces, it is not going to be true that $f^\sharp$ is surjective if and only if $f_x^\sharp$ is surjective for all $x\in X$. To make this clearer, let's assume $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$, so $f=\mathrm{Spec}(\alpha)$ for $\alpha:A\to B$ a ring homomorphism. The stalk map of $f$ at $x=\mathfrak{q}\in\mathrm{Spec}(B)$ is the ring map $A_\mathfrak{p}\to B_\mathfrak{q}$, where $\mathfrak{q}=\alpha^{-1}(\mathfrak{p})$. In general, surjectivity of this map for all $\mathfrak{q}\in\mathrm{Spec}(B)$ does not imply surjectivity of $\alpha$ itself.
I think the simplest example that will illustrate this is when $B=A_g$ is a principal localization of $A$. Then in fact the stalk map in the previous paragraph is an isomorphism for every prime ideal of $A_g$ (the set of which are in natural bijection with the set of primes of $A$ not containing $g$, i.e. $D(g)$). But the localization map $A\to A_g$ (i.e. the map on global sections of $f$) is not usually surjective. Note that in this case $f$ is a homeomorphism onto the open subset $D(g)$ of $\mathrm{Spec}(A)$, but $D(g)$ is not generally closed in $A$.
I think maybe this illustrates why the first condition is important, and why, if one wants to think about surjectivity of $f^\sharp$ in terms of the stalks of $f$, $f_x^\sharp$, for $x\in X$, the topological condition is needed, and logically "precedes" the condition on $f^\sharp$.
Lastly, I should note that the maps which I have been calling the ``stalks of $f$," $f_x^\sharp$, for $x\in X$, are in fact the stalks of the map of sheaves on $X$ (in the usual sense) $f^\flat:f^{-1}\mathscr{O}_Y\to \mathscr{O}_X$ corresponding to $f^\sharp$ under the adjunction between $f^{-1}$ and $f_*$. So surjectivity of all $f_x^\sharp$, $x\in X$, is logically equivalent to surjectivity of $f^\flat$. There is no reason to believe that $f^\flat$ is surjective if and only if $f^\sharp$ is, or even that there is an implication in either direction in general.
Best Answer
Usually it is just supposed to mean that the underlying map on the topological spaces is surjective.
There are various other possibilities for ”surjective-like“ morphisms (e.g. epi, dominant etc.), but they have other names.
There are non-surjective epis and surjective non-epis in the category of schemes. Thus do not arbitrarily switch between these two.