hcju/stacksqa huggingface dataset:
Let $V_1$ and $V_2$ be dense subsets of $T_1'$ and $T'_2$, respectively, that are open in $U_1$ and $U_2$ respectively (see proof of Lemma \ref{lemma-map-of-components}). The images of $|V_1|$ and $|V_2|$ in $|\mathcal{X}|$ are non-empty open subsets of the irreducible subset $T$, and therefore have non-empty intersection. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian}, the map $|V_1 \times_\mathcal{X} V_2| \to |V_1| \times_{|\mathcal{X}|} |V_2|$ is surjective. Consequently $V_1 \times_\mathcal{X} V_2$ is a non-empty algebraic space; we may therefore choose an \'etale surjection $V \to V_1 \times_\mathcal{X} V_2$ whose source is a (non-empty) scheme. If we let $T'$ be any irreducible component of $V$, then Lemma \ref{lemma-map-of-components} shows that the closure of the image of $T'$ in $U_1$ (respectively $U_2$) is equal to $T'_1$ (respectively $T'_2$). \medskip\noindent Applying Lemma \ref{lemma-multiplicities} twice we find that $$ m_{T_1', U_1} = m_{T', V} = m_{T_2', U_2}, $$ as required.
Let $A$ be as in Definition \ref{definition-formal-branches} and let $\Spec(A) \to \mathcal{X}$ be as in Remark \ref{remark-upgrade}. We claim that the generic point of an irreducible component of $\Spec(A)$ maps to a generic point of an irreducible component of $|\mathcal{X}|$. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Consider the diagram $$ \xymatrix{ \Spec(A) \times_\mathcal{X} U \ar[d]_p \ar[r]_-q & U \ar[d]^f \\ \Spec(A) \ar[r]^j & \mathcal{X} } $$ By Lemma \ref{lemma-versal-ring-flat} we see that $j$ is flat. Hence $q$ is flat. On the other hand, $f$ is surjective smooth hence $p$ is surjective smooth. This implies that any generic point $\eta \in \Spec(A)$ of an irreducible component is the image of a codimension $0$ point $\eta'$ of the algebraic space $\Spec(A) \times_\mathcal{X} U$ (see Properties of Spaces, Section \ref{spaces-properties-section-generic-points} for notation and use going down on \'etale local rings). Since $q$ is flat, $q(\eta')$ is a codimension $0$ point of $U$ (same argument). Since $U$ is a scheme, $q(\eta')$ is the generic point of an irreducible component of $U$. Thus the closure of the image of $q(\eta')$ in $|\mathcal{X}|$ is an irreducible component by Lemma \ref{lemma-map-of-components} as claimed. \medskip\noindent Clearly the claim provides a mechanism for defining the desired map. To see that it is surjective, we choose $u_0 \in U$ mapping to $x_0$ in $|\mathcal{X}|$. Choose an affine open $U' \subset U$ neighbourhood of $u_0$. After shrinking $U'$ we may assume every irreducible component of $U'$ passes through $u_0$. Then we may replace $\mathcal{X}$ by the open substack corresponding to the image of $|U'| \to |\mathcal{X}|$. Thus we may assume $U$ is affine has a point $u_0$ mapping to $x_0 \in |\mathcal{X}|$ and every irreducible component of $U$ passes through $u_0$. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian} there is a point $t \in |\Spec(A) \times_\mathcal{X} U|$ mapping to the closed point of $\Spec(A)$ and to $u_0$. Using going down for the flat local ring homomorphisms $$ A \longrightarrow \mathcal{O}_{\Spec(A) \times_\mathcal{X} U, \overline{t}} \longleftarrow \mathcal{O}_{U, u_0} $$ we see that every minimal prime of $\mathcal{O}_{U, u_0}$ is the image of a minimal prime of the local ring in the middle and such a minimal prime maps to a minimal prime of $A$. This proves the surjectivity. Some details omitted.
By Lemma \ref{lemma-Artin-approximation-by-smooth-morphism} we may assume there is a smooth morphism $U \to \mathcal{X}$ where $U$ is a scheme and a $k$-valued point $u_0$ of $U$ such that $\mathcal{O}_{U, u_0}^\wedge$ is a versal ring to $\mathcal{X}$ at $x_0$. By construction of our map in the proof of Lemma \ref{lemma-branches} (which simplifies greatly because $A = \mathcal{O}_{U, u_0}^\wedge$) we find that it suffices to show: the multiplicity of an irreducible component of $U$ passing through $u_0$ is the same as the multiplicity of any irreducible component of $\Spec(\mathcal{O}_{U, u_0}^\wedge)$ mapping into it. \medskip\noindent Translated into commutative algebra we find the following: Let $C = \mathcal{O}_{U, u_0}$. This is essentially of finite type over $\mathcal{O}_{S, s}$ and hence is a G-ring (More on Algebra, Proposition \ref{more-algebra-proposition-finite-type-over-G-ring}). Then $A = C^\wedge$. Therefore $C \to A$ is a regular ring map. Let $\mathfrak q \subset C$ be a minimal prime and let $\mathfrak p \subset A$ be a minimal prime lying over $\mathfrak q$. Then $$ R = C_\mathfrak p \longrightarrow A_\mathfrak p = R' $$ is a regular ring map of Artinian local rings. For such a ring map it is always the case that $$ \text{length}_R R = \text{length}_{R'} R' $$ This is what we have to show because the left hand side is the multiplicity of our component on $U$ and the right hand side is the multiplicity of our component on $\Spec(A)$. To see the equality, first we use that $$ \text{length}_R(R) \text{length}_{R'}(R'/\mathfrak m_R R') = \text{length}_{R'}(R') $$ by Algebra, Lemma \ref{algebra-lemma-pullback-module}. Thus it suffices to show $\mathfrak m_R R' = \mathfrak m_{R'}$, which is a consequence of being a regular homomorphism of zero dimensional local rings.
Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms} can be used to verify that the right hand side $\dim_u(U) + \dim_u(U_x)$ is independent of the choice of the smooth morphism $U \to \mathcal{X}$ and $u \in |U|$. We omit the details. In particular, we may assume $U$ is a scheme. In this case we can compute $\dim_u(U_x)$ by choosing the representative of $x$ to be the composite $\Spec \kappa(u) \to U \to \mathcal{X}$, where the first morphism is the canonical one with image $u \in U$. Then, if we write $R = U \times_{\mathcal{X}} U$, and let $e : U \to R$ denote the diagonal morphism, the invariance of relative dimension under base-change shows that $\dim_u(U_x) = \dim_{e(u)}(R_u)$. Thus we see that the right hand side is equal to $\dim_u (U) - \dim_{e(u)}(R_u) = \dim_x(\mathcal{X})$ as desired.
Choose a smooth surjective morphism $V \to \mathcal{U}$ whose source is a scheme, and let $v\in |V|$ be a point mapping to $u$. Then the composite $V \to \mathcal{U} \to \mathcal{X}$ is also smooth, and by Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms} we have $\dim_x(\mathcal{X}) = \dim_v(V) - \dim_v(V_x),$ while $\dim_u(\mathcal{U}) = \dim_v(V) - \dim_v(V_u).$ Thus $$ \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) = \dim_v (V_x) - \dim_v (V_u). $$ \medskip\noindent Choose a representative $\Spec k \to \mathcal{X}$ of $x$ and choose a point $v' \in | V \times_{\mathcal{X}} \Spec k|$ lying over $v$, with image $u'$ in $|\mathcal{U}\times_{\mathcal{X}} \Spec k|$; then by definition $\dim_u(\mathcal{U}_x) = \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \Spec k),$ and $\dim_v(V_x) = \dim_{v'}(V\times_{\mathcal{X}} \Spec k).$ \medskip\noindent Now $V\times_{\mathcal{X}} \Spec k \to \mathcal{U}\times_{\mathcal{X}}\Spec k$ is a smooth surjective morphism (being the base-change of such a morphism) whose source is an algebraic space (since $V$ and $\Spec k$ are schemes, and $\mathcal{X}$ is an algebraic stack).  Thus, again by definition, we have \begin{align*} \dim_{u'}(\mathcal{U}\times_{\mathcal{X}} \Spec k) & = \dim_{v'}(V\times_{\mathcal{X}} \Spec k) - \dim_{v'}(V \times_{\mathcal{X}} \Spec k)_{u'}) \\ & = \dim_v(V_x) - \dim_{v'}( (V\times_{\mathcal{X}} \Spec k)_{u'}). \end{align*} Now $V\times_{\mathcal{X}} \Spec k \cong V\times_{\mathcal{U}} (\mathcal{U}\times_{\mathcal{X}} \Spec k),$ and so Lemma \ref{lemma-base-change-invariance-of-relative-dimension} shows that $\dim_{v'}((V\times_{\mathcal{X}} \Spec k)_{u'})  = \dim_v(V_u).$ Putting everything together, we find that $$ \dim_u(\mathcal{U}) - \dim_x(\mathcal{X}) = \dim_u(\mathcal{U}_x), $$ as required.
Suppose to begin with that $\mathcal{T}$ is a scheme $T$, let $U \to \mathcal{X}$ be a smooth surjective morphism whose source is a scheme, and let $T' = T \times_{\mathcal{X}} U$. Let $f': T' \to U$ be the pull-back of $f$ over $U$, and let $g: T' \to T$ be the projection. \medskip\noindent Lemma \ref{lemma-base-change-invariance-of-relative-dimension} shows that $\dim_{t'}(T'_{f'(t')}) = \dim_{g(t')}(T_{f(g(t'))}),$ for $t' \in T'$, while, since $g$ is smooth and surjective (being the base-change of a smooth surjective morphism) the map induced by $g$ on underlying topological spaces is continuous and open (by Properties of Spaces, Lemma \ref{spaces-properties-lemma-topology-points}), and surjective. Thus it suffices to note that part (1) for the morphism $f'$ follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-openness-bounded-dimension-fibres}, and part (2) from either of Morphisms, Lemma \ref{morphisms-lemma-flat-finite-presentation-CM-fibres-relative-dimension} or Morphisms, Lemma \ref{morphisms-lemma-smooth-omega-finite-locally-free} (each of which gives the result for schemes, from which the analogous results for algebraic spaces can be deduced exactly as in Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-openness-bounded-dimension-fibres}. \medskip\noindent Now return to the general case, and choose a smooth surjective morphism $h:V \to \mathcal{T}$ whose source is a scheme. If $v \in V$, then, essentially by definition, we have $$ \dim_{h(v)}(\mathcal{T}_{f(h(v))}) = \dim_{v}(V_{f(h(v))}) - \dim_{v}(V_{h(v)}). $$ Since $V$ is a scheme, we have proved that the first of the terms on the right hand side of this equality is upper semi-continuous (and even locally constant if $f$ is smooth), while the second term is in fact locally constant. Thus their difference is upper semi-continuous (and locally constant if $f$ is smooth), and hence the function $\dim_{h(v)}(\mathcal{T}_{f(h(v))})$ is upper semi-continuous on $|V|$ (and locally constant if $f$ is smooth). Since the morphism $|V| \to |\mathcal{T}|$ is open and surjective, the lemma follows.
The equivalence of the two claims follows directly from the definitions.   Suppose, then, that $U\subset X$ is a non-empty open subset. Certainly $\dim U \leq \dim X$, and we have to show that $\dim U \geq \dim X.$ Write $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say $$ X = Z_0 \supset Z_1 \supset \ldots \supset Z_d. $$ Since $X$ is Jacobson, the minimal irreducible closed subset $Z_d$ is equal to $\{x\}$ for some closed point $x$. \medskip\noindent If $x \in U,$ then $$ U = U \cap Z_0  \supset U\cap Z_1 \supset \ldots \supset U\cap Z_d = \{x\} $$ is a strictly decreasing chain of irreducible closed subsets of $U$, and so we conclude that $\dim U \geq d$, as required.  Thus we may suppose that $x \not\in U.$ \medskip\noindent Consider the flat morphism $\Spec \mathcal{O}_{X,x} \to X$. The non-empty (and hence dense) open subset $U$ of $X$ pulls back to an open subset $V \subset \Spec \mathcal{O}_{X,x}$. Replacing $U$ by a non-empty quasi-compact, and hence Noetherian, open subset, we may assume that the inclusion $U \to X$ is a quasi-compact morphism.  Since the formation of scheme-theoretic images of quasi-compact morphisms commutes with flat base-change Morphisms, Lemma \ref{morphisms-lemma-flat-base-change-scheme-theoretic-image} we see that $V$ is dense in $\Spec \mathcal{O}_{X,x}$, and so in particular non-empty, and of course $x \not\in V.$  (Here we use $x$ also to denote the closed point of $\Spec \mathcal{O}_{X,x}$, since its image is equal to the given point $x \in X$.) Now $\Spec \mathcal{O}_{X,x} \setminus \{x\}$ is Jacobson Properties, Lemma \ref{properties-lemma-complement-closed-point-Jacobson} and hence $V$ contains a closed point $z$ of $\Spec \mathcal{O}_{X,x} \setminus \{x\}$.  The closure in $X$ of the image of $z$ is then an irreducible closed subset $Z$ of $X$ containing $x$, whose intersection with $U$ is non-empty, and for which there is no irreducible closed subset properly contained in $Z$ and properly containing $\{x\}$ (because pull-back to $\Spec \mathcal{O}_{X,x}$ induces a bijection between irreducible closed subsets of $X$ containing $x$ and irreducible closed subsets of $\Spec \mathcal{O}_{X,x}$). Since $U \cap Z$ is a non-empty closed subset of $U$, it contains a point $u$ that is closed in $X$ (since $X$ is Jacobson), and since $U\cap Z$ is a non-empty (and hence dense) open subset of the irreducible set $Z$ (which contains a point not lying in $U$, namely $x$), the inclusion $\{u\} \subset U\cap Z$ is proper. \medskip\noindent As $X$ is catenary, the chain $$ X = Z_0 \supset Z \supset \{x\} = Z_d $$ can be refined to a chain of length $d+1$, which must then be of the form $$ X = Z_0 \supset W_1 \supset \ldots \supset W_{d-1} = Z \supset \{x\} = Z_d. $$ Since $U\cap Z$ is non-empty, we then find that $$ U = U \cap Z_0 \supset U \cap W_1\supset \ldots \supset U\cap W_{d-1} = U\cap Z \supset \{u\} $$ is a strictly decreasing chain of irreducible closed subsets of $U$ of length $d+1$, showing that $\dim U \geq d$, as required.
By assumption we may find a smooth surjective morphism $U \to \mathcal{X}$ whose source is a universally catenary scheme. The base-change $U\times_{\mathcal{X}} \mathcal{Y}$ is then an algebraic stack; let $V \to U\times_{\mathcal{X}} \mathcal{Y}$ be a smooth surjective morphism whose source is a scheme. The composite $V \to U\times_{\mathcal{X}} \mathcal{Y} \to \mathcal{Y}$ is then smooth and surjective (being a composite of smooth and surjective morphisms), while the morphism $V \to U\times_{\mathcal{X}} \mathcal{Y} \to U$ is locally of finite type (being a composite of morphisms that are locally finite type).  Since $U$ is universally catenary, we see that $V$ is universally catenary (by Morphisms, Lemma \ref{morphisms-lemma-universally-catenary-local}), as claimed.
It suffices to show that $\dim_x(\mathcal{X})$ is locally constant on $|\mathcal{X}|$, since it will then necessarily be constant (as $|\mathcal{X}|$ is connected, being irreducible).  Since $\mathcal{X}$ is pseudo-catenary, we may find a smooth surjective morphism $U \to \mathcal{X}$ with $U$ being a universally catenary scheme.  If $\{U_i\}$ is an cover of $U$ by quasi-compact open subschemes, we may replace $U$ by $\coprod U_i,$, and it suffices to show that the function $u \mapsto \dim_{f(u)}(\mathcal{X})$ is locally constant on $U_i$. Since we check this for one $U_i$ at a time, we now drop the subscript, and write simply $U$ rather than $U_i$. Since $U$ is quasi-compact, it is the union of a finite number of irreducible components, say $T_1 \cup \ldots \cup T_n$.  Note that each $T_i$ is Jacobson, catenary, and locally Noetherian, being a closed subscheme of the Jacobson, catenary, and locally Noetherian scheme $U$. \medskip\noindent By Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms}, we have $\dim_{f(u)}(\mathcal{X}) = \dim_{u}(U) - \dim_{u}(U_{f(u)}).$ Lemma \ref{lemma-relative-dimension-is-semi-continuous} (2) shows that the second term in the right hand expression is locally constant on $U$, as $f$ is smooth, and hence we must show that $\dim_u(U)$ is locally constant on $U$.  Since $\dim_u(U)$ is the maximum of the dimensions $\dim_u T_i$, as $T_i$ ranges over the components of $U$ containing $u$, it suffices to show that if a point $u$ lies on two distinct components, say $T_i$ and $T_j$ (with $i \neq j$), then $\dim_u T_i = \dim_u T_j$, and then to note that $t\mapsto \dim_t T$ is a constant function on an irreducible Jacobson, catenary, and locally Noetherian scheme $T$ (as follows from Lemma \ref{lemma-constancy-of-dimension}). \medskip\noindent Let $V = T_i \setminus (\bigcup_{i' \neq i} T_{i'})$ and $W = T_j \setminus (\bigcup_{i' \neq j} T_{i'})$. Then each of $V$ and $W$ is a non-empty open subset of $U$, and so each has non-empty open image in $|\mathcal{X}|$.  As $|\mathcal{X}|$ is irreducible, these two non-empty open subsets of $|\mathcal{X}|$ have a non-empty intersection. Let $x$ be a point lying in this intersection, and let $v \in V$ and $w\in W$ be points mapping to $x$. We then find that $$ \dim T_i = \dim V = \dim_v (U) = \dim_x (\mathcal{X}) + \dim_v (U_x) $$ and similarly that $$ \dim T_j = \dim W = \dim_w (U) = \dim_x (\mathcal{X}) + \dim_w (U_x). $$ Since $u \mapsto \dim_u (U_{f(u)})$ is locally constant on $U$, and since $T_i \cup T_j$ is connected (being the union of two irreducible, hence connected, sets that have non-empty intersection), we see that $\dim_v (U_x) = \dim_w(U_x)$, and hence, comparing the preceding two equations, that $\dim T_i = \dim T_j$, as required.
Choose a smooth surjective morphism $U\to \mathcal{X}$ whose source is a scheme; the base-changed morphism $V = U\times_{\mathcal{X}} \mathcal{Z} \to \mathcal{Z}$ is then also smooth and surjective, and the projection $V \to U$ is a closed immersion. If $v \in |V|$ maps to $z \in |\mathcal{Z}|$, and if we let $u$ denote the image of $v$ in $|U|$, then clearly $\dim_v(V) \leq \dim_u(U)$, while $\dim_v (V_z) = \dim_u(U_x)$, by Lemma \ref{lemma-base-change-invariance-of-relative-dimension}. Thus $$ \dim_z(\mathcal{Z})  = \dim_v(V) - \dim_v(V_z) \leq \dim_u(U) - \dim_u(U_x) = \dim_x(\mathcal{X}), $$ as claimed.
Lemma \ref{lemma-closed-immersions} shows that $\dim_x (\mathcal{T}) \leq \dim_x(\mathcal{X})$ for each irreducible component $\mathcal{T}$ passing through the point $x$.   Thus to prove the lemma, it suffices to show that \begin{equation} \label{equation-desired-inequality} \dim_x(\mathcal{X}) \leq \sup_{\mathcal{T}} \{\dim_x(\mathcal{T})\}. \end{equation} Let $U\to\mathcal{X}$ be a smooth cover by a scheme. If $T$ is an irreducible component of $U$ then we let $\mathcal{T}$ denote the closure of its image in $\mathcal{X}$, which is an irreducible component of $\mathcal{X}$. Let $u \in U$ be a point mapping to $x$. Then we have $\dim_x(\mathcal{X})=\dim_uU-\dim_uU_x=\sup_T\dim_uT-\dim_uU_x$, where the supremum is over the irreducible components of $U$ passing through $u$. Choose a component $T$ for which the supremum is achieved, and note that $\dim_x(\mathcal{T})=\dim_uT-\dim_u T_x$. The desired inequality (\ref{equation-desired-inequality}) now follows from the evident inequality $\dim_u T_x \leq \dim_u U_x.$ (Note that if $\Spec k \to \mathcal{X}$ is a representative of $x$, then $T\times_{\mathcal{X}} \Spec k$ is a closed subspace of $U\times_{\mathcal{X}} \Spec k$.)
Choose a smooth surjective morphism  $f:U \to \mathcal{X}$ whose source is a scheme, and consider the function $u \mapsto \dim_{f(u)}(\mathcal{X});$ since the morphism $|U| \to |\mathcal{X}|$ induced by $f$ is open (as $f$ is smooth) as well as  surjective (by assumption), and takes finite type points to finite type points (by the very definition of the finite type points of $|\mathcal{X}|$), it suffices to show that for any $u \in U$, and any open neighbourhood of $u$, there is a finite type point $u_0$ in this neighbourhood such that $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$ Since, with this reformulation of the problem,  the surjectivity of $f$ is no longer required, we may replace $U$ by the open neighbourhood of the point $u$ in question, and thus reduce to the problem of showing that for each $u \in U$, there is a finite type point $u_0 \in U$ such that $\dim_{f(u_0)}(\mathcal{X}) = \dim_{f(u)}(\mathcal{X}).$ By Lemma \ref{lemma-behaviour-of-dimensions-wrt-smooth-morphisms} $\dim_{f(u)}(\mathcal{X}) = \dim_u(U) - \dim_u(U_{f(u)}),$ while $\dim_{f(u_0)}(\mathcal{X}) = \dim_{u_0}(U) - \dim_{u_0}(U_{f(u_0)}).$ Since $f$ is smooth, the expression $\dim_{u_0}(U_{f(u_0)})$ is locally constant as $u_0$ varies over $U$ (by Lemma \ref{lemma-relative-dimension-is-semi-continuous} (2)), and so shrinking $U$ further around $u$ if necessary, we may assume it is constant.  Thus the problem becomes to show that we may find a finite type point $u_0 \in U$ for which $\dim_{u_0}(U) = \dim_u(U)$. Since by definition $\dim_u U$ is the minimum of the dimensions $\dim V$, as $V$ ranges over the open neighbourhoods $V$ of $u$ in $U$, we may shrink $U$ down further around $u$ so that $\dim_u U = \dim U$. The existence of desired point $u_0$ then follows from Lemma \ref{lemma-dimension-achieved-by-finite-type-point}.
This is true because $s : \mathcal{O}_{U, \overline{u}} \to \mathcal{O}_{R, e(\overline{u})}$ is a flat local homomorphism of Noetherian local rings and hence $$ \dim(\mathcal{O}_{R, e(\overline{u})}) = \dim(\mathcal{O}_{U, \overline{u}}) + \dim(\mathcal{O}_{R_u, e(\overline{u})}) $$ by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}.
Suppose we have two smooth neighbourhoods $(U, u)$ and $(U', u')$ of $x$ with $u$ and $u'$ finite type points. After shrinking $U$ and $U'$ we may assume that $u$ and $u'$ are closed points (by definition of finite type points). Then we choose a surjective \'etale morphism $W \to U \times_\mathcal{X} U'$. Let $W_u$ be the fibre of $W \to U$ over $u$ and let $W_{u'}$ be the fibre of $W \to U'$ over $u'$. Since $u$ and $u'$ map to the same point of $|\mathcal{X}|$ we see that $W_u \cap W_{u'}$ is nonempty. Hence we may choose a closed point $w \in W$ mapping to both $u$ and $u'$. This reduces us to the discussion in the next paragraph. \medskip\noindent Assume $(U', u') \to (U, u)$ is a smooth morphism of smooth neightbourhoods of $x$ with $u$ and $u'$ closed points. Goal: prove the invariant defined for $(U, u)$ is the same as the invariant defined for $(U', u')$. To see this observe that $\mathcal{O}_{U, u} \to \mathcal{O}_{U', u'}$ is a flat local homomorphism of Noetherian local rings and hence $$ \dim(\mathcal{O}_{U', \overline{u}'}) = \dim(\mathcal{O}_{U, \overline{u}}) + \dim(\mathcal{O}_{U'_u, \overline{u}'}) $$ by Algebra, Lemma \ref{algebra-lemma-dimension-base-fibre-equals-total}. (We omit working through all the steps to relate properties of local rings and their strict henselizations, see More on Algebra, Section \ref{more-algebra-section-permanence-henselization}). On the other hand we have $$ R' = U' \times_{U, t} R \times_{s, U} U' $$ Thus we see that $$ \dim(\mathcal{O}_{R', e(\overline{u}')}) = \dim(\mathcal{O}_{R, e(\overline{u})}) + \dim(\mathcal{O}_{U'_u \times_u U'_u, (\overline{u}', \overline{u}')}) $$ To prove the lemma it suffices to show that $$ \dim(\mathcal{O}_{U'_u \times_u U'_u, (\overline{u}', \overline{u}')}) = 2\dim(\mathcal{O}_{U'_u, \overline{u}'}) $$ Observe that this isn't always true (example: if $U'_u$ is a curve and $u'$ is the generic point of this curve). However, we know that $u'$ is a closed point of the algebraic space $U'_u$ locally of finite type over $u$. In this case the equality holds because, first $\dim_{(u', u')}(U'_u \times_u U'_u) = 2\dim_{u'}(U'_u)$ by Varieties, Lemma \ref{varieties-lemma-dimension-product-locally-algebraic} and second the agreement of dimension with dimension of local rings in closed points of locally algebraic schemes, see Varieties, Lemma \ref{varieties-lemma-dimension-locally-algebraic}. We omit the translation of these results for schemes into the language of algebraic spaces.