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Let $I$ be an arbitrary index set and consider the category $\bf{Sets}$. For an object $X$, show that $X^{I}$ satisfies the universal property with respect to the constant family $X_i = X$ for all $i \in I$, that is, $$X^I \cong \prod_{i \in I} X_i$$

How can I show this? I have troubles to understand the set $X^I$, whats the definition of a set raised to the power of some index set?

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    $\begingroup$ Usually $X^I$ denotes the set of all functions $I\to X$. Edit: In this case you have to show that there is a bijective correspondence between the set of all functions $I\to X$, and the set of all families $(x_i)_{i\in I}$ in $X$ (i.e., a family indexed by $I$ with elements in $X$). Think about the definition of a sequence! $\endgroup$ – Nesta Oct 5 '16 at 15:43
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$X^I$ denotes the set of all functions $I\to X$.
Note that we can regard such a function $f\in X^I$ as a collection of the elements $(f(i))_{i\in I}$ indexed by $I$.

For every $i\in I$, let $p_i:=X^I\to X$ the map $f\mapsto f(i)$. These will be the projections.

The universal property states that, for any set $Y$ and any collection $u_i:Y\to X$ indexed by $I$, there exists a unique $\bar u:Y\to X^I$ such that $p_i\circ \bar u=u_i$ for each $i\in I$.

When writing out uniqueness, you will find how to define $\bar u$ for given $u_i$'s, so existence will be easy.

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