Diagonal of immersion is an isomorphism

algebraic-geometryschemes

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.

Best Answer

  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.

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