Diagonal of immersion is an isomorphism

Question

Let $f:X \rightarrow Y$ be a morphism of schemes. We say that $f$ is an immersion if it can be decomposed as $f=g \circ h$, where $h$ is a closed immersion and $g$ an open immersion. Then, the diagonal morphism $\Delta_{X/Y}: X \rightarrow X \times_{Y} X$ is an isomorphism.

My approach has been to show that when $f$ is an open/closed immersion then the diagonal is an isomorphism, and then to relate the diagonals of $g$ and $h$ to the diagonal of $f$ in some way, but I have not succeeded in this.

Does someone know how to prove this result? Thanks in advance.

Answer 1

1. In any category with pullbacks, if $f\colon X\rightarrow Y$ is a morphism and $\Delta\colon X\rightarrow X\times_YX$ the associated diagonal, then $\Delta$ is an isomorphism if and only $f$ is a monomorphism. (Exercise in category theory.)

2. The composition of monomorphisms is a monomorphism.

3. Open immersions are monomorphisms. Indeed, let $(i,i^{\#})\colon(U,\mathcal{O}_U)\rightarrow(X,\mathcal{O}_X)$ be an open immersion and $(f,f^{\#}),(g,g^{\#})\colon(T,\mathcal{O}_T)\rightarrow(U,\mathcal{O}_U)$ be two morphisms such that $(i,i^{\#})\circ(f,f^{\#})=(i,i^{\#})\circ(g,g^{\#})$. In particular, $i\circ f=i\circ g$, hence $f=g$. Furthermore, for all open $V\subseteq U$, the composites $\mathcal{O}_X(V)=\mathcal{O}_U(V)\stackrel{f^{\#}_V}{\rightarrow}\mathcal{O}_T(f^{-1}(V))$ and $\mathcal{O}_X(V)=\mathcal{O}_U(V)\stackrel{g^{\#}_V}{\rightarrow}\mathcal{O}_T(g^{-1}(V))$ agree, hence $f^{\#}=g^{\#}$.

4. Closed immersions are monomorphisms. Indeed, let $(i,i^{\#})\colon(Y,\mathcal{O}_Y)\rightarrow(X,\mathcal{O}_X)$ be a closed immersion and $(f,f^{\#}),(g,g^{\#})\colon(T,\mathcal{O}_T)\rightarrow(Y,\mathcal{O}_Y)$ be two morphisms such that $(i,i^{\#})\circ(f,f^{\#})=(i,i^{\#})\circ(g,g^{\#})$. Then, as in the previous step, we obtain $f=g$. Now, $i_{\ast}f^{\#}\circ i^{\#}=i_{\ast}g^{\#}\circ i^{\#}$ as morphisms $\mathcal{O}_X\rightarrow i_{\ast}\mathcal{O}_Y\rightarrow i_{\ast}f_{\ast}\mathcal{O}_T=i_{\ast}g_{\ast}\mathcal{O}_T$, but $i^{\#}\colon\mathcal{O}_X\rightarrow i_{\ast}\mathcal{O}_Y$ is an epimorphisms of sheaves (this is either by definition or requires an argument), hence $i_{\ast}f^{\#}=i_{\ast}g^{\#}$ as morphisms $i_{\ast}\mathcal{O}_Y\rightarrow i_{\ast}f_{\ast}\mathcal{O}_T=i_{\ast}g_{\ast}\mathcal{O}_T$. However, $i$ being a topological embedding implies that $i_{\ast}\colon\mathbf{Sh}(Y)\rightarrow\mathbf{Sh}(X)$ is a faithful functor (exercise), hence $f^{\#}=g^{\#}$.

The desired claim follows by putting these together. Note that none of this really has anything to do with schemes and it works just as well in the category of locally ringed spaces.

Answer 2

I did this differently than Thorgott, but that seems like a nice solution. Honestly, I have never proven that open/closed immersions are monomorphisms so I'd be interested in seeing the details.

Anyway, here is how I did this problem when in came up in Vakil's notes.

First, let's understand how close $\Delta: X \to X \times_Y X$ is to being a closed immersion. For each $x \in X$, and affine neighborhood $V$ of $f(x)$, we can find an affine neighborhood $U \subset f^{-1}(V)$ of $x$, so that $\Delta|_U: U \to U\times_V U$ is a closed immersion. We also notice that by construction, $U = \Delta^{-1}(U \times_V U). $ (let me know if this isn't clear.) As such, $\Delta$ can be viewed as a morphism $g: S \to T$ satisfying the following property:

For every $s \in S$, there is a neighborhood $V_s$ of $g(s)$ so that $g|_{g^{-1}(V)}: g^{-1}(V) \to V$ is a closed immersion.

Now, let $V = \bigcup_{s \in S} V_s \subset T$. Then, $g(S) \subset V$, so $g$ factors $$S \stackrel{g'}\to V \hookrightarrow T$$

and $V \hookrightarrow T$ is an open immersion. As such, we just need to show that $g'$ is a closed immersion. But now, this just follows from the local nature of closed immersions, which can be found here and here.