\(x\in {d(h(c))}\) if and only if \(h(c)\notin {x}\), which is true if and only if \(c\notin {f(x)}\) (because \(f=Spec(h)\)), which is equivalent to \(f(x)\in {d(c)}\).

We first show that \(z(h(b))\subseteq {z(h(a))}\). Indeed, if \(x\in {X}\) belongs to \(z(h(b))\), then by Lemma 3 we have that \(f(x)\in {z(b)}\). As \(z(b)\subseteq {z(a)}\), \(f(x)\in {z(a)}\). Then Lemma 3 implies that \(x\in {z(h(a))}\).

By Theorem 3 of Hochster’s paper, we now only need to show that for any element \(p=(y,x)\) of \(\sigma (X)\) such that \(h(a)(y)\neq {0}\) it is true that \(v_p(h(b))\leq {v_p(h(a))}\), with equality only if \(h(b)(x)\neq {0}\).

Indeed, if \(y\in {d(h(a))}\), then by Lemma 3 we know that \(f(y)\in {d(a)}\). Because \((f(y),f(x))\in \sigma (X')\), Theorem 3 in Hochster’s thesis implies that \(v_{f(p)}(a)\geq {v}_{f(p)}(b)\). By the definition of an indexed spring morphism, \(v_{f(p)}(a)=v_p(h(a))\) and \(v_{f(p)}(b)=v_p(h(b))\). Hence \(v_p(h(a))\geq {v_p(h(b))}\).

If \(v_p(h(b))={v_p(h(a))}\), then \(v_{f(p)}(a)={v}_{f(p)}(b)\) and so \(f(x)\in {d(b)}\). Thus, by Lemma 3 we have that \(x\in {d(h(b))}\) and we are done!

The most difficult part of the proof is to show that if \(r=q(a{\# }b)\in {A’}\) where \(q=\Sigma _{i=0}^m{a_i}t^i\), then \(r':=\Sigma _{i=0}^m{h(a_i)}(h(a){\# }h(b))^i\) belongs to \(A\) and equals \(h(r)\). The purpose of showing this is to guarantee that if we construct the homomorphism \(h_1\) in the obvious way, then \(h_1\) is well-defined.

Let \(c:=b^m{r}\). Then \(h(b^m)r'=\Sigma _{i=0}^m{h(a_i)}h(a)^i{h(b)^{m-i}}=h(b^m)h(r)\). We then try to prove that \(r'(x)=h(r)(x)\) for any \(x\in {X}\). We show this by classifying the \(x\)’s. If \(x\in {d(h(b))}\), then it is clear that \(r'(x)=h(r)(x)\) because we have shown that \(h(b^m)r'=h(b^m)h(r)\).

Now assume \(x\in {z(h(b))}\). Let \(\overline{h}:\frac{A'}{h^{-1}(x)}\rightarrow \frac{A}{x}\) be the ring homomorphism induced by \(h\). Then
\(r'(x)=h(a_0)(x)=h(a_0)+x(\in \frac{A}{x})=\overline{h}(a_0+h^{-1}(x))\)
\(=\overline{h}(a_0(f(x)))=\overline{h}(r(f(x)))=\overline{h}(r+h^{-1}(x))\)
\(=h(r)+x(\in \frac{A}{x})=h(r)(x)\).

It is easy to show that the obvious way to define \(h_1\) satisfy the axioms of ring homomorphisms, and that this extension of \(h\) is unique.

Let \(r=q(a\# b)\) where \(q=\Sigma _{i=0}^{m}{a_i}t^i\in {A’}[t]\). We first show that \(f^{-1}(z(r))=z(h_1(r))\). By a previous comment, this is sufficient for proving that \((f,h_1)\) is a spring morphism from \(\boldsymbol {A}[h(a)\# h(b)]\) to \(\boldsymbol {A’}[a\# b]\), as \(r\) has been taken arbitrarily.

Pick some \(x\in {X}\). Let \(c:=b^m{r}=\Sigma _{i=0}^m{a_i}a^i{b^{m-i}}\). Then \(h(c)=\Sigma _{i=0}^m{h(a_i)}{h(a)^i}h(b)^{m-i}=h(b)^m{h_1(r)}\) and we have:
\(x\in {f^{-1}}(z(r))\)
\(\Leftrightarrow {f(x)\in {z(r)}}=(z(c)\cap {d}(a))\cup (z(a_0)\cap {z}(a))\)
\(\Leftrightarrow {x}\in (z(h(c))\cap {d}(h(a)))\cup (z(h(a_0))\cap {z}(h(a)))\)
\(\Leftrightarrow {x}\in {z(h_1(r))}\).

We still need to show that for any \(p=(y,x)\in \sigma (X)\) with \(y\in {d}(h_1(r))\) it is true that \(v_p(h_1(r))=v_{f(p)}(r)\). Indeed, because \(h(c)=h(b)^m{h_1}(r)\), \(v_p(h_1(r))=v_p(h(c))-v_p(h(b^m))=v_{f(p)}(c)-v_{f(p)}(b^m)=v_{f(p)}(r)\).

We show that for any \(x\in {X}\), \(h_2(a\# b)(x)=(h(a)\# h(b))(x)\). Indeed, if \(h(b)(x)\neq {0}\), then as
\(h_2(a\# b)h(b)=h_2(a\# b)h_2(b)=h_2(a)=h(a)=(h(a)\# h(b))h(b)\),
we immediately have that \(h_2(a\# b)(x)=(h(a)\# h(b))(x)\). If \(h(b)(x)=0\), then \(x\in {z}(h_2(b))\) and so \(f(x)\in {z}(b)\) by Lemma 3. This means \((a\# b)(f(x))=0\). Then Lemma 3 implies \(h_2(a\# b)(x)=0\), so \(h_2(a\# b)(x)=(h(a)\# h(b))(x)\).