[proofplan] We prove the Lipschitz inequalities directly from the formula for the [product metric](/page/Product%20Metric). For two arbitrary points of $X\times Y$, the finite-$p$ formula shows that each coordinate distance is bounded above by the $p$-sum of the two coordinate distances. The case $p=\infty$ is the same comparison with the maximum. These coordinate inequalities are exactly the definition that $\pi_X$ and $\pi_Y$ are $1$-Lipschitz. [/proofplan] [step:Compare each coordinate distance with the finite product distance] Assume first that $1 \le p < \infty$. Let $(x_1,y_1),(x_2,y_2)\in X\times Y$. Define \begin{align*} a=d_X(x_1,x_2) \end{align*} and \begin{align*} b=d_Y(y_1,y_2). \end{align*} Since $d_X$ and $d_Y$ are metrics, $a\ge 0$ and $b\ge 0$. The definition of $d_p$ gives \begin{align*} d_p((x_1,y_1),(x_2,y_2))=(a^p+b^p)^{1/p}. \end{align*} Because $a^p \le a^p+b^p$ and the function $t\mapsto t^{1/p}$ is increasing on $[0,\infty)$, we obtain \begin{align*} a\le (a^p+b^p)^{1/p}=d_p((x_1,y_1),(x_2,y_2)). \end{align*} Similarly, since $b^p\le a^p+b^p$, \begin{align*} b\le (a^p+b^p)^{1/p}=d_p((x_1,y_1),(x_2,y_2)). \end{align*} Therefore \begin{align*} d_X(x_1,x_2)\le d_p((x_1,y_1),(x_2,y_2)) \end{align*} and \begin{align*} d_Y(y_1,y_2)\le d_p((x_1,y_1),(x_2,y_2)). \end{align*} [guided] Assume $1 \le p < \infty$, and choose arbitrary points $(x_1,y_1),(x_2,y_2)\in X\times Y$. The goal is to compare the distance between the first coordinates and the distance between the second coordinates with the product distance between the ordered pairs. Define the two coordinate distances by \begin{align*} a=d_X(x_1,x_2) \end{align*} and \begin{align*} b=d_Y(y_1,y_2). \end{align*} Because $d_X$ and $d_Y$ are metrics, both $a$ and $b$ are nonnegative [real numbers](/page/Real%20Numbers). The finite-$p$ product metric is built from these two numbers by \begin{align*} d_p((x_1,y_1),(x_2,y_2))=(a^p+b^p)^{1/p}. \end{align*} Why does this dominate each coordinate distance? Since $b^p\ge 0$, we have $a^p\le a^p+b^p$. The map $t\mapsto t^{1/p}$ is increasing on $[0,\infty)$, so taking $p$th roots preserves the inequality: \begin{align*} a\le (a^p+b^p)^{1/p}. \end{align*} Substituting back the definitions of $a$ and $d_p$, this becomes \begin{align*} d_X(x_1,x_2)\le d_p((x_1,y_1),(x_2,y_2)). \end{align*} The second coordinate is identical except that we use $a^p\ge 0$ instead of $b^p\ge 0$. Thus $b^p\le a^p+b^p$, and again monotonicity of $t\mapsto t^{1/p}$ gives \begin{align*} b\le (a^p+b^p)^{1/p}. \end{align*} Substituting back, \begin{align*} d_Y(y_1,y_2)\le d_p((x_1,y_1),(x_2,y_2)). \end{align*} So, for finite $p$, the product distance always bounds both coordinate distances from above. [/guided] [/step] [step:Compare each coordinate distance with the supremum product distance] Assume now that $p=\infty$. Let $(x_1,y_1),(x_2,y_2)\in X\times Y$, and define $a=d_X(x_1,x_2)$ and $b=d_Y(y_1,y_2)$. By definition, \begin{align*} d_\infty((x_1,y_1),(x_2,y_2))=\max\{a,b\}. \end{align*} The maximum of two real numbers is greater than or equal to each of them, so \begin{align*} d_X(x_1,x_2)\le d_\infty((x_1,y_1),(x_2,y_2)) \end{align*} and \begin{align*} d_Y(y_1,y_2)\le d_\infty((x_1,y_1),(x_2,y_2)). \end{align*} [/step] [step:Apply the definition of a $1$-Lipschitz map to the coordinate projections] Let $(x_1,y_1),(x_2,y_2)\in X\times Y$ be arbitrary. The coordinate projections are the maps \begin{align*} \pi_X:X\times Y &\to X \end{align*} and \begin{align*} \pi_Y:X\times Y &\to Y \end{align*} defined by $\pi_X(x,y)=x$ and $\pi_Y(x,y)=y$. Hence \begin{align*} d_X(\pi_X(x_1,y_1),\pi_X(x_2,y_2))=d_X(x_1,x_2) \end{align*} and \begin{align*} d_Y(\pi_Y(x_1,y_1),\pi_Y(x_2,y_2))=d_Y(y_1,y_2). \end{align*} From the finite-$p$ comparison when $1\le p<\infty$, and from the supremum comparison when $p=\infty$, both inequalities \begin{align*} d_X(\pi_X(x_1,y_1),\pi_X(x_2,y_2))\le d_p((x_1,y_1),(x_2,y_2)) \end{align*} and \begin{align*} d_Y(\pi_Y(x_1,y_1),\pi_Y(x_2,y_2))\le d_p((x_1,y_1),(x_2,y_2)) \end{align*} hold for every pair of points in $X\times Y$. This is precisely the definition that $\pi_X$ and $\pi_Y$ are $1$-Lipschitz maps from $(X\times Y,d_p)$ to $(X,d_X)$ and $(Y,d_Y)$, respectively. [/step]