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Does the characteristic property for products of smooth manifolds hold as well:
$$f\text{ smooth}\iff \pi_i\circ f\text{ smooth ...where }f:M\to\prod_{i\in I} M_i\text{ and }\pi_i:\prod_{i\in I} M_i\to M_i$$

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The answer is: YES!!! =D

Before we start off we define charts on finite(!) products of manifolds in the obvious way (obvious!)...

Now, observe that the composition of differentiable functions is differentiable (tricky!).
This immediately implies the analogue for manifolds.
Besides, note that projection functions are differentiable (easy!); so its analogue for manifolds.

These two observations imply that the components of differentiable functions are differentiable; mutatis mutandis for manifolds.

Conversely, it follows that differentiable components of finite(!) products give rise to differentiable functions (doable!); this again immediately implies the analogue for manifolds.

Concluding that a function is differentiable iff its components are differentiable; mutatis mutandis for manifolds.

Especially, applying this to the original problem the answer is 'yes':
The product of manifolds satisfyies the characteristic property for smooth manifolds.

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