I'm reading about category theory and the term 'induce' is kind of mysterious to me. What does it mean for a particular map to induce another map? For instance, the following example was given:
$$ \require{AMScd} \begin{CD} X @>>> X^I\\ @V{f}VV @V{f_*}VV \\ Y @>>> Y^I \end{CD}$$
(I might be drawing it badly) But basically, there's a correspondence between the category of sets and the category of cartesian products of sets that associates a set $X$ and and it's cartesian product $X^I$. Then, if we have a map $f : X \to Y$ between two sets, this map will induce a map $f_*$ that acts on $X^I$ in a 'canonical' way, such that:
$$f_* ((x_i)_{i \in I}) := (f(x_i))_{i \in I}$$
it seems fine to just acknowledge this as a 'canonical' map as the map that is induced, but is there any way to actually formalise what it means for one map to induce another? What sets the map $f_*$ apart from the others?
This is usually the result of some universal property being satisfied. Here are some examples:
Universal property of quotient groups: For any group homomorphism $\phi:G\to H$ and any subgroup $K\subseteq\ker\phi$, there exists a unique homomorphism $\bar{\phi}:G/K\to H$ such that $\phi = \bar{\phi}\circ\pi$, where $\pi:G\to G/K$ is the natural projection. So, if we have a map $\phi:G\to H$ and a subgroup $K\subseteq \ker\phi$, then we say that $\phi:G\to H$ induces the map $\bar{\phi}:G/K\to H$.
Universal property of products: If $f:Z\to X$ and $g:Z\to Y$ are two maps, then there exists a unique map $f\times g:Z\to X\times Y$ such that $\pi_X\circ (f\times g) = f$ and $\pi_Y\circ(f\times g) = g$. Here, $\pi_X:X\times Y\to X$ and $\pi_Y:X\times Y\to Y $ are the obvious projections. So, we would say that the map $f\times g$ is induced by the maps $f$ and $g$.
Unless I'm mistaken these are examples of what Hagen von Itzen and Arthur are explaining. You can read more about these here and here.
It can also come from the application of a functor, (see here) which is your example. Let $F:\textbf{Set}\to\textbf{Set}$ be defined by $$F(X\xrightarrow{\ f\ }Y) = X^I\xrightarrow{\ f_* \ }Y^I.$$ In this case $F(f) = f_*$ is your induced map.