What does it mean to induce a map?

Question

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?

Answer 1

I think the examples are being a bit too technical. "Induce" isn't a term; it's largely just being used as a normal word. For example, you may also see the phrase "gives rise to" which Google gives as the very definition of "induce". If I was going to give a more technical definition, I'd say it's an arrow (or other construction) that is functionally determined from some (often unclearly) specified data.

Your example corresponds to a functorial action which is literally a family of functions^*. Given $X$, $Y$, and $f$, $f_*$ is functionally determined by the action of the functor $(-)^I : \mathbf{Set}\to\mathbf{Set}^I$.

A universal property corresponds to a representable functor, that is a functor $F : \mathcal{C}^{op}\to\mathbf{Set}$ which is naturally isomorphic to $\mathcal{C}(-,X)$ for some given $X$. This is again literally a family of functions. Given a $Y$ in $\mathcal{C}$ and an element of $FY$ an arrow from $Y \to X$ is uniquely determined.

We could say that the natural isomorphism $F\cong\mathcal{C}(-,X)$ gives rise to or induces the function $FY \to \mathcal{C}(Y,X)$. This is just an instance of the whole determining a part, e.g. if I have an element of $A\times B$ then clearly I also have an element of $A$ uniquely specified (via projection) from the element I started with. While not usually articulated, this is not that uncommon. In particular, it's often the case that we're mostly interested in only a component of a limit cone. A case that you've probably already run into (or you will shortly) is a change of base situation via pullback. For example, given a family of sets $\{X_i\}_{i\in I}$ indexed by $I$, we have a map $\{X_i\}_{i\in I} \to I$. Given further a map $\sigma : J \to I$ we can create a new family of maps, $\{X_{\sigma(j)}\}_{j\in J} \to J$, indexed by $J$ by pulling back along $\sigma$. This is often presented as a functor (induced by $\sigma$) $\sigma^* : \mathbf{Set}/I \to \mathbf{Set}/J$ called a pullback functor. Of course, this ignores that the pullback also gives us an arrow $\{X_{\sigma(j)}\}_{j\in J} \to \{X_i\}_{i\in I}$.

Of course, we can also have an explicit construction (which will likely correspond to a functorial action or universal property though there's no need to mention that). For example, given a linear transformation $f : U \to V$, we can make an outermorphism, $\bar f$ via $\bar f(u_1\wedge\dots\wedge u_n) = f(u_1)\wedge\dots\wedge f(u_n)$ which is a linear transformation between exterior algebras $\Lambda U \to \Lambda V$. This is an example of a functorial action, but there is no need to establish that to say that $f$ induces $\bar f$.

As a final scenario, we may have a diagram for which only one arrow in a certain spot will make it commute. This is common for universal properties, but it can happen in a situation that doesn't correspond to a universal property. In this case we're essentially "solving for" an arrow.

There are other possibilities, e.g. a representative of an equivalence class determines the equivalence class. For example, a path in a topological space induces an arrow in the fundamental groupoid. My point is that there is no specific notion being referred to by "induces" or "gives rise to". If an author intends there to be, they will define these as terms.

^* at least between locally small categories

Answer 2

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.

Answer 3

In language form, we say an induced map arises from the composition map by restricting one of its components. Moreover, it is found that every continuous function gives an induced map.