3
$\begingroup$

I am working through Category for Scientists and am having trouble with the following question:

Let $A$ and $B$ be the sets defined by $A := \{a, 7, Q\}$ and $B := \{r8, "Bob", \clubsuit\}$. Note that the sets $A$ and $B$ are isomorphic. Supposing that $f : B \rightarrow \{1, 2, 3, 4, 5\}$ sends "Bob" to 1, sends $\clubsuit$ to 3, and sends $r8$ to 4, is there a canonical function $A \rightarrow \{1, 2, 3, 4, 5\}$ corresponding to $f$?

I'm not entirely certain what the question is asking. What do "canonical function" and "corresponding to $f$" mean?

$\endgroup$
2
  • 1
    $\begingroup$ Its basically saying: okay, there are many functions $A \rightarrow B$, is there any such function that deserves to be considered special? If there was such a special entity $g : A \rightarrow B$, then from the function $f : B \rightarrow (*)$ given, we can obtain a new function $f' : A \rightarrow (*)$ in a "canonical" way, by defining $f' = f \circ g$. However, since there is really no entity $g : A \rightarrow B$ that is in any way special relative to the others, thus there is no "canonical" way obtaining a function $f' : A \rightarrow (*)$ from the given function $f : B \rightarrow (*)$. $\endgroup$ Dec 30, 2013 at 14:51
  • $\begingroup$ By the way, this is why category theorists we tend not to identify isomorphic objects. On the other hand, if we're given objects $A$ and $B$ such that there exists a unique isomorphism $f : A \rightarrow B$, then it is sensible to identify $A$ and $B$. $\endgroup$ Dec 30, 2013 at 14:54

2 Answers 2

5
$\begingroup$

The point of the question is to highlight the fact that though $A$ and $B$ are isomorphic, there is no "obvious" or "canonical" or "natural" isomorphism between them.

If you have a function $f : B \to \{1,2,3,4,5\}$, you can "translate" it to a function $A \to \{1,2,3,4,5\}$ by composing it with a suitable bijection (isomorphism) between $A$ and $B$. So the answer is no, from the previous paragraph.

$\endgroup$
1
  • $\begingroup$ +1 Apparently I misread the question to mean "find a canonical function". Thank you for the clarification. $\endgroup$
    – Code-Guru
    Dec 30, 2013 at 20:58
0
$\begingroup$

Just wanted to add that you should also look at the note at the bottom of the 21st page:

Canonical means something like “best choice”, a choice that stands out as the only reasonable one.

$\endgroup$
2
  • $\begingroup$ Thanks. It has been a long time since I posted this. If I recall correctly, my confusion was because there is no obvious "best choice" among the 6 possible bijections from $A$ to $B$. $\endgroup$
    – Code-Guru
    Dec 26, 2017 at 15:40
  • 2
    $\begingroup$ Your question is still relevant! You must be a category theory master by now but other people who read this book might be confused by the term. $\endgroup$
    – Danny Mor
    Dec 27, 2017 at 8:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .