Add a lemma suggested by Matthew Emerton

See http://stacks.math.columbia.edu/tag/04YY#comment-785

stacks-morphisms.tex

@@ -1300,6 +1300,36 @@ \section{Higher diagonals}
 as discussed above, this proves the lemma.
 \end{proof}
 
+\noindent
+This lemma leads to the following hierarchy for
+morphisms of algebraic stacks.
+
+\begin{lemma}
+\label{lemma-hierarchy}
+A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is
+\begin{enumerate}
+\item a monomorphism if and only if $\Delta_{f, 1}$ is an isomorphism,
+\item representable by algebraic spaces if and only if $\Delta_{f, 1}$
+is a monomorphism,
+\item the second diagonal $\Delta_{f, 2}$ is always a monomorphism.
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+Recall from Properties of Stacks, Lemma
+\ref{stacks-properties-lemma-monomorphism}
+that a morphism of algebraic stacks is a monomorphism
+if and only if its diagonal is an isomorphism of stacks.
+Thus Lemma \ref{lemma-second-diagonal}
+can be rephrased as saying that a morphism is
+representable by algebraic spaces if the diagonal
+is a monomorphism. In particular, it shows that condition
+(3) of Lemma \ref{lemma-properties-diagonal-representable}
+is actually an if and only if, i.e., a morphism of algebraic stacks
+is representable by algebraic spaces if and only if
+its diagonal is a monomorphism.
+\end{proof}
+
 \begin{lemma}
 \label{lemma-first-diagonal-separated-second-diagonal-closed}
 Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.