A useful observation is that $\kappa(s) = \kappa(t)$ by the existence of $\sigma$. Observe that the statement makes sense as the fibres of $Y_\sigma$ and $Y_{\sigma'}$ over $s \in \Spec(\mathcal{O}_{S, s}^\wedge)$ are both canonically isomorphic to $Y_t$. We will think of the property ``$\sigma'$ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$'' as a constraint on $\sigma'$. If we have several such constraints, say given by $(T'_i, t'_i) \to (T, t)$, $i = 1, \ldots, n$ then we can combined them by considering $(T'_1 \times_T \ldots \times_T T'_n, (t'_1, \ldots, t'_n)) \to (T, t)$. We will use this without further mention in the following. \medskip\noindent By Lemma \ref{lemma-control-agreement} we can assume that any $\sigma'$ as in the statement of the lemma is the same as $\sigma$ modulo $\mathfrak m_s^2$. Note that the conormal algebra of $Y_t$ in $Y_\sigma$ is just the quasi-coherent graded $\mathcal{O}_{Y_t}$-algebra $$ \bigoplus\nolimits_{n \geq 0} \mathfrak m_s^n\mathcal{O}_{Y_\sigma}/ \mathfrak m_s^{n + 1}\mathcal{O}_{Y_\sigma} $$ and similarly for $Y_{\sigma'}$. Since $\sigma$ and $\sigma'$ agree modulo $\mathfrak m_s^2$ we see that these two algebras are the same in degrees $0$ and $1$. On the other hand, these conormal algebras are generated in degree $1$ over degree $0$. Hence if there is an isomorphism extending the isomorphism just constructed in degrees $0$ and $1$, then it is unique. \medskip\noindent We may assume $S$ and $T$ are affine. Let $Y = Y_1 \cup \ldots \cup Y_n$ be an affine open covering. If we can construct $(T_i', t'_i) \to (T, t)$ as in the lemma such that the desired isomorphism (see previous paragraph) exists for $Y_i \to T \to S$ and $\sigma$, then these glue by uniqueness to prove the result for $Y \to T$. Thus we may assume $Y$ is affine. \medskip\noindent Write $S = \Spec(R)$, $T = \Spec(C)$, and $Y = \Spec(B)$. Choose a presentation $B = C[x_1, \ldots, x_n]/(f_1, \ldots, f_m)$. Denote $R^\wedge = \mathcal{O}_{S, s}^\wedge$. Let $a_{kj} \in R^\wedge[x_1, \ldots, x_n]$ be polynomials such that $$ \sum\nolimits_{j = 1, \ldots, m} a_{kj}\sigma(f_j) = 0,\quad \text{for }k = 1, \ldots, K $$ is a set of generators for the module of relations among the $\sigma(f_j) \in R^\wedge[x_1, \ldots, x_n]$. Thus we have an exact sequence \begin{equation} \label{equation-resolution} R^\wedge[x_1, \ldots, x_n]^{\oplus K} \to R^\wedge[x_1, \ldots, x_n]^{\oplus m} \to R^\wedge[x_1, \ldots, x_n] \to B \otimes_{C, \sigma} R^\wedge \to 0 \end{equation} Let $c$ be an integer which works in the Artin-Rees lemma for both the first and the second map in this sequence and the ideal $\mathfrak m_{R^\wedge}R^\wedge[x_1, \ldots, x_n]$ as defined in More on Algebra, Section \ref{more-algebra-section-artin-rees}. Write $$ a_{kj} = \sum\nolimits_{I \in \Omega} a_{kj, I} x^I \quad\text{and}\quad f_j = \sum\nolimits_{I \in \Omega} f_{j, I} x^I $$ in multiindex notation where $a_{kj, I} \in R^\wedge$, $f_{j, I} \in C$, and $\Omega$ a finite set of multiindices. Then we see that $$ \sum\nolimits_{j = 1, \ldots, m,\ I, I' \in \Omega,\ I + I' = I''} a_{kj, I} \sigma(f_{j, I'}) = 0,\quad I''\text{ a multiindex} $$ in $R^\wedge$. Thus we take $$ C' = C[t_{jk, I}]/ \left( \sum\nolimits_{j = 1, \ldots, m,\ I, I' \in \Omega,\ I + I' = I''} t_{kj, I} f_{j, I'},\ I''\text{ a multiindex}\right) $$ Then $\sigma$ factors through a map $\tilde\sigma : C' \to R^\wedge$ sending $t_{kj, I}$ to $a_{jk, I}$. Thus $T' = \Spec(C')$ comes with a point $t' \in T'$ such that $\sigma$ factors through $\mathcal{O}_{T, t} \to \mathcal{O}_{T', t'}$. Let $t_{kj} = \sum t_{kj, I} x^I$ in $C'[x_1, \ldots, x_n]$. Then we see that we have a complex \begin{equation} \label{equation-resolution-new} C'[x_1, \ldots, x_n]^{\oplus K} \to C'[x_1, \ldots, x_n]^{\oplus m} \to C'[x_1, \ldots, x_n] \to B \otimes_C C' \to 0 \end{equation} which is exact at $C'[x_1, \ldots, x_n]$ and whose base change by $\tilde\sigma$ gives (\ref{equation-resolution}). \medskip\noindent By Lemma \ref{lemma-control-agreement} we can find a further morphism $(T'', t'') \to (T', t')$ such that $\tilde\sigma$ factors through $\mathcal{O}_{T', t'} \to \mathcal{O}_{T'', t''}$ and such that if $\sigma' : C \to R^\wedge$ factors through $\mathcal{O}_{T'', t''}$, then the induced map $\tilde \sigma' : C' \to R^\wedge$ agrees modulo $\mathfrak m_s^{c + 1}$ with $\tilde \sigma$. Thus if $\sigma'$ is such a map, then we obtain a complex $$ R^\wedge[x_1, \ldots, x_n]^{\oplus K} \to R^\wedge[x_1, \ldots, x_n]^{\oplus m} \to R^\wedge[x_1, \ldots, x_n] \to B \otimes_{C, \sigma'} R^\wedge \to 0 $$ over $R^\wedge[x_1, \ldots, x_n]$ by applying $\tilde\sigma'$ to the polynomials $t_{kj}$ and $f_j$. In other words, this is the base change of the complex (\ref{equation-resolution-new}) by $\tilde\sigma'$. The matrices defining this complex are congruent modulo $\mathfrak m_s^{c + 1}$ to the matrices defining the complex (\ref{equation-resolution}) because $\tilde \sigma$ and $\tilde \sigma'$ are congruent modulo $\mathfrak m_s^{c + 1}$. Since (\ref{equation-resolution}) is exact, we can apply More on Algebra, Lemma \ref{more-algebra-lemma-approximate-complex-graded} to conclude that $$ \text{Gr}_{\mathfrak m_s}(B \otimes_{C, \sigma'} R^\wedge) \cong \text{Gr}_{\mathfrak m_s}(B \otimes_{C, \sigma} R^\wedge) $$ as desired.