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If you consider the map $x \mapsto (x, \phi(x))$ and call it $g$$\Phi $, then you have that $h = f \circ g$$h = f \circ \Phi$.
If you consider the map $x \mapsto (x, \phi(x))$ and call it $g$, then you have that $h = f \circ g$.
If you consider the map $x \mapsto (x, \phi(x))$ and call it $\Phi $, then you have that $h = f \circ \Phi$.
