Theorem $:$ Suppose $(X,d_X)$ is a metric space, $(Y,d_Y)$ is a complete metric space, $f : X \times Y \longrightarrow Y,$ $\exists K \lt 1$ such that $d_Y (f(x,y), f(x,y')) \leq K d_Y (y,y').$ Then there exists unique continuous function $\varphi : X \longrightarrow Y$ such that $\varphi (x) = f(x, \varphi (x)),$ for all $x \in X.$
Proof $:$ For each $x \in X,$ $f_x : Y \longrightarrow Y$ be defined by $f_x (y) = f(x,y),\ y \in Y.$ Then by the given condition it follows that $f_x$ is a contraction map for each $x \in X.$ Hence it follows from Banach principle of contraction that $f_x$ has a unique fixed point for each $x \in X$ (Since $Y$ is complete). This induces an implicit function $\varphi : X \longrightarrow Y$ such that $\varphi (x) = f(x, \varphi (x)),$ for all $x \in X.$ So it only remains to show that $\varphi$ is continuous on $X.$ For that we define $g_x : X \longrightarrow Y$ by $g_x (u) = f(u, \varphi (x)), u \in X.$ Then $g_x$ is continuous on $X$ for each $x \in X$ since it is the composition of the continuous map $u \longmapsto (u, \varphi (x))$ with $f.$ In particular $g_x$ is continuous at $x$ for each $x \in X.$ So for a given $\varepsilon \gt 0$ there exists a $\delta \gt 0$ such that for all $z \in X$ with $d_X (z,x) \lt \delta$ we have $$d_Y (g_x (z), g_x(x)) \lt (1-K) \varepsilon$$ or in other words for all $z \in X$ with $d_X (z,x) \lt \delta$ we have $$d_Y (f(z, \varphi (x)), f(x, \varphi (x))) \lt (1-K) \varepsilon.$$ Therefore for all $z \in X$ with $d_X (z,x) \lt \delta$ we have \begin{align*} d_Y (\varphi (x), \varphi (z)) & \leq d_Y(f (x, \varphi (x)), f(z, \varphi (x))) + d_Y (f(z, \varphi (x)), f (z, \varphi (z))) \\ & \lt (1-K) \varepsilon + Kd_Y (\varphi (x), \varphi (z)) \end{align*} This shows that for all $z \in X$ with $d_X (z,x) \lt \delta$ we have $$d_Y (\varphi (x), \varphi (z)) \lt \varepsilon,$$ proving that $\varphi$ is continuous at $x.$ Since $x \in X$ was arbitrarily taken it follows that $\varphi$ is continuous on $X.$
This completes the proof.
QED
I have understood the above proof. What I think is that the above proof is independent of the choice of the metric on $X \times Y.$ Because in the entire proof the metric on $X \times Y$ is of no use. But our instructor says that $X \times Y$ is endowed with product metric. I can't agree with him. Am I missing something? Any help in this regard will be appreciated.
Thanks in advance.