We formulate our main result in this section which is a version for into linear isometries of C(X,τ)-spaces of a known Holsztyńskis’s theorem (Theorem 2) for into linear isometries of C𝔽(X)-spaces. We first study unit-preserving linear isometries from C(X,τ) into C(Y,η).

Proof.
Let E=T(C(X,τ)). Since T:C(X,τ)→C(Y,η) is a linear isometry, we deduce that E is a uniformly closed linear subspace of C(Y,η). Moreover, T(1X)=1Y implies that 1Y∈E. We define the map S:C(X,τ)→E by Sf=Tf. Then, S is a linear isometry from the real Banach space (C(X,τ),∥·∥X) onto the real Banach space (E,∥·∥Y). Therefore, S* is a linear isometry from E* onto C(X,τ)* and so
(42)S*(Ext(BE*))=Ext(BC(X,τ)*).
Now we define
(43)Y0={y∈Y:S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*)}.
We first show that Y0 is nonempty. To prove this fact, we show that the following statement, namely (I), holds.

(I) For each x∈X there exists y0∈Y such that S*(φC(Y,η),y,1|E)=φC(X,τ),x,1.

Let x∈X. Then φC(X,τ),x,1∈Ext(BC(X,τ)*) by Theorem 1. From (42), there exists Λ∈Ext(BE*) such that
(44)S*Λ=φC(X,τ),x,1.
Since E is a linear subspace of C(Y,η), there exists Λ~∈Ext(BC(Y,η)*) such that
(45)Λ~|E=Λ.
By Theorem 1, there exists (y0,γ0)∈Y×𝕊ℂ such that
(46)Λ~=φC(X),y0,γ0.
From (44), (45), and (46), we have
(47)S*(φC(Y,η),y0,γ0|E)=φC(X,τ),x,1.
Since 1X∈C(X,τ) and T1X=1Y, by (47) we have
(48)Re(γ0)=Re(γ01Y(y0))=φC(Y,η),y0,γ0(1Y)=φC(Y,η),y0,γ0(T1X)=φC(Y,η),y0,γ0(S1X)=S*(φC(Y,η),y0,γ0|E)(1X)=φC(X,τ),x,1(1X)=Re(1eC(X),x(1X))=Re(1X(x))=Re(1)=1.
This implies that γ0=1 since |γ0|=1. Hence, by (47) we have
(49)S*(φC(Y,η),y0,1|E)=φC(X,τ),x,1,
and so statement (I) holds.

Since φC(X,τ),x,1∈Ext(BC(X,τ)*) for all x in X, statement (I) implies that Y0 is nonempty.

We next show that the following statement, namely, (II), holds.

(II) For each y∈Y0 there exists a unique point x∈X such that (Tf)(y)=f(x) for all f in C(X,τ).

Let y∈Y0. Then, S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*). Hence, by Theorem 1, there exists (x0,λ0)∈X×𝕊ℂ such that
(50)S*(φC(Y,η),y,1|E)=φC(X,τ),x0,λ0.
Since 1X∈C(X,τ) and T1X=1Y, by (50) we have
(51)Re(λ0)=Re(λ0·1X(x0))=φC(X,τ),x0,λ0(1X)=S*(φC(Y,η),y,1|E)(1X)=φC(Y,η),y,1(S1X)=φC(Y,η),y,1(T1X)=φC(Y,η),y,1(1Y)=Re(1eC(Y,η),y(1Y)) =Re(eC(Y,η),y(1Y))=Re(1Y(y))=Re(1)=1.
This implies that λ0=1 since |λ0|=1. Thus, by (50) we have
(52)S*(φC(Y,η),y,1|E)=φC(X,τ),x0,1.
From (52) we give
(53)Re((Sf)(y))=Re(f(x0)), ∀f∈C(X,τ).
It follows that
(54)Re((Tf)(y))=Re(f(x0)), ∀f∈C(X,τ).

According to (54) and condition (ii), E is extremely regular at {y,η(y)}. On the other hand, C(X,τ) is extremely regular at {x0,τ(x0)} by Lemma 4. Hence, by Lemma 3, we deduce that (Sf)(y)=f(x0) for all f in C(X,τ) or (Sf)(y)=f(τ(x0)) for all f in C(X,τ). We assume that x=x0 whenever (Sf)(y)=f(x0) for all f in C(X,τ) and x=τ(x0) whenever (Sf)(y)=f(τ(x0)) for all f in C(X,τ). Then, x∈X and we have
(55)(Tf)(y)=f(x), ∀f∈C(X,τ).
This proves the existent of x. To show uniqueness, assume that there exists x′∈X such that
(56)(Tf)(y)=f(x′), ∀f∈C(X,τ).
From (55) and (56) we have f(x)=f(x′) for all f in C(X,τ). This implies that x=x′ since C(X,τ) separates the points of X. Thus, statement (II) holds.

Now we define the map h:Y0→X by h(y)=x whenever (Tf)(y)=f(x) for all f in C(X,τ). The statement (II) implies that h is well-defined. By definition of h, we have
(57)(Tf)(y)=f(h(y)), ∀f∈C(X,τ) ∀y∈Y0.
We prove that h∘η=τ∘h in Y0. Let y∈Y0. For each f∈C(X,τ), by (57) we have
(58)f((h∘η)(y))=f(h(η(y)))=(Tf)(η(y))=(Tf)(y)¯=f(h(y))¯=f(τ(h(y)))=f((τ∘h)(y)).
This implies that (h∘η)(y)=(τ∘h)(y) since C(X,τ) separates the points of X. Hence, h∘η=τ∘h on Y0.

Continually, we show that Y0 is a η-invariant. Let y∈Y0. For each f∈C(X,τ) we have
(59)S*(φC(Y,η),η(y),1|E)(f)=φC(Y,η),η(y),1(Sf)=φC(Y,η),η(y),1(Tf)=Re(1eC(Y),η(y)),1(Tf)) =Re((Tf)(η(y)))=Re((Tf)(y)¯)=Re((f(h(y)))¯)=Re(1f(τ(h(y))))= Re(1eC(X),τ(h(y))|C(X,τ)(f)) =(Re(1eC(X),τ(h(y))|C(X,τ)))(f) =φC(X,τ),τ(h(y)),1(f).
This implies that
(60)S*(φC(Y,η),η(y),1|E)=φC(X,τ),τ(h(y)),1.
Hence, S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*) by Theorem 1. Therefore, η(y)∈Y0 and so Y0 is η-invariant.

Now we show that h is surjective. Let x∈X. According to statement (I), we deduce that there exists a point y0∈Y0 such that
(61)S*(φC(Y,η),y0,1)=φC(X,τ),τ(x),1.
It follows that y0∈Y and we have
(62)Re((Tf)(y0))=Re(f(x)), ∀f∈C(X,τ).
Statement (II) implies that there exists a point x′∈X such that
(63)(Tf)(y0)=f(x′), ∀f∈C(X,τ).
From (62) and (63), we have Re(f(x))=Re(f(x′)) for all f in C(X,τ). This implies that x′∈{x,τ(x)} since Re C(X,τ) separates the points of X/τ. If x′=x, then by (63) and definition of h we have x=h(y0). If x′=τ(x), then by (63) and definition of h we have
(64)(Tf)(η(y0))=(Tf)(y0)¯=f(x′)¯=f(τ(x′))=f(x),
for all f in C(X,τ), and so x=h(η(y0)). Therefore, h is surjective.

Continually, we show that Y0 is a boundary for E. Let g∈E. Then, there exists a function f∈C(X,τ) such that Tf=g. Since f is a continuous complex-valued function on the compact space X, there exists a point x0∈X such that ∥f∥X=|f(x0)|. The surjectivity of h:Y0→X implies that there exists a point y0∈Y0 such that h(y0)=x0. Hence, (Tf)(y0)=f(x0), and so
(65)∥g∥Y=∥Tf∥Y=∥f∥X=|f(x0)|=|(Tf)(y0)|=|g(y0)|.
Therefore, Y0 is a boundary for E.

We now check that h:Y0→X is continuous. Let y∈Y0, and let {yα}α∈J be a net in Y0 such that limαyα=y in Y0. Then, for each f∈C(X,τ) we have limα(Tf)(yα)=(Tf)(y) in X. This implies that for each limαf(h(yα))=f(h(y)) in X. Since C(X,τ) separates the points of X, we have limαh(yα)=h(y) in X. Therefore, h:Y0→X is continuous.

Finally, we show that Y0 is closed in Y. Let y∈Y0¯, the closure of Y0 in Y. Then, there exists a net {yα}α∈J in Y0 such that limαyα=y in Y. Then, for each f∈C(X,τ) we have limα(Tf)(yα)=(Tf)(y), and so limαf(h(yα))=(Tf)(y) in X. Since the net {f(h(yα))}α∈J converges for all f in C(X,τ), we conclude that the net {h(yα)}α∈J converges in X to a point x∈X. Hence, for each f∈C(X,τ) we have limαf(h(yα))=f(x). Therefore, (Tf)(y)=f(h(y)) for all f in C(X,τ). This implies that for each f∈C(X,τ) we have
(66)S*(φC(Y,η),y,1|E)(f)=(φC(Y,η),y,1|E)(Sf)=(φC(Y,η),y,1)(Tf)=Re(1eC(Y),y(Tf))= Re((Tf)(y))= Re(f(h(y)))=Re(eC(X),h(y)(f))=(Re(1eC(X),h(y)|C(X,τ)))(f)=φC(X,τ),h(y),1(f).
Hence,
(67)S*(φC(Y,η),y,1|E)=φC(X,τ),h(y),1.
This implies that S*(φC(Y,η),y,1|E)∈Ext(BC(X,τ)*), by Theorem 1. Therefore, y∈Y0. So Y0 is closed in Y.

We now study the into case (not necessarily unit-preserving).

Proof.
Assume that a=T1X. Then, a∈C(Y,η), and by (i) we have |a(y)|=1 for all y∈Y. Clearly, (a¯)(Tf)∈C(Y,η) for all C(Y,η), and (a¯)(T1X)=1Y. We define the map a¯·T:C(X,τ)→C(Y,η) by (a¯·T)(f)=(a¯)(Tf). It is easy to see that a¯·T is a linear isometry from C(X,τ) into C(Y,η) such that (a¯·T)(1X)=1Y.

Applying (ii), we can easily show that if y∈Y and there exists a point x∈X such that Re(((a¯·T)f)(y))=Re(f(x)) for all f in C(X,τ), then (a¯·T)(C(X,τ)) is extremely regular at {y,η(y)}. Let E=T(C(X,τ)). Then, a¯E=(a¯·T)(C(X,τ)). We define S:C(X,τ)→E by Sf=Tf and S1:C(X,τ)→a¯E by S1f=(a¯·T)f. Clearly, S is a linear isometry from C(X,τ) onto E, and S1 is a linear isometry from C(X,τ) onto a¯E. Now, we define
(69)Y0={y∈Y:S*(φC(Y,η),y,a(y)¯|E)∈Ext(BC(X,τ)*)}.
It is easy to see that
(70)Y0={y∈Y:S1*(φC(Y,η),y,1|a¯E)∈Ext(BC(X,τ)*)}.
By given arguments in the proof of Theorem 10, Y0 is a η-invariant closed boundary for a¯E, and there exists a continuous function h from Y0 onto X with h∘η=τ∘h such that
(71)((a¯·T)(f))(y)=f(h(y)), ∀f∈C(X,τ), ∀y∈Y0.
We now show that Y0 is a boundary for E. Let g∈E. Then, a¯g∈a¯E. Hence, there exists a point y0∈Y0 such that
(72)∥a¯g∥Y=|(a¯g)(y0)|.
Since |a(y)|=1 for all y∈Y, we conclude that
(73)∥a¯g∥Y=∥g∥Y.
From (72) and (73), we have
(74)∥g∥Y=|a¯g(y0)|=|a¯(y0)g(y0)|=|a(y0)¯g(y0)|=|a(y0)¯||g(y0)|=|a(y0)||g(y0)|=|g(y0)|.
Therefore, Y0 is a boundary for E.

Finally, we show that
(75)(Tf)(y)=a(y)f(h(y)), ∀f∈C(X,τ), ∀y∈Y0.
Let f∈C(X,τ) and y∈Y0. From (73), we have
(76)((a¯·T)(f))(y)=f(h(y)).
Applying (76), we deduce that
(77)(Tf)(y)=a(y)a(y0)¯(Tf)(y)=a(y)((a¯)(Tf))(y)=a(y)((a¯·T)f)(y)=a(y)f(h(y)).
Hence, the proof is complete.