I have a nit-picky question about how the word "object" (as in "mathematical object") is generally used/understood. I'll ask by way of a simple, specific example.

Consider 1) the set of permutations of $n$ things, and 2) the set of bijective functions mapping $n$ things to $n$ things. If you separately explain 1) and 2) to someone with no prior knowledge, it's not immediately obvious that they are basically the same thing (otherwise I've chosen a bad example).

Would you say,

  • 1) and 2) are two equivalent objects
  • 1) and 2) are two different ways of looking at the same object


  • both the above usages are ok, depending on what you want to emphasize
  • $\begingroup$ Presumably the three "$n$ things" can be three different (but fixed) objects? Also, you should perhaps specify which of the (equivalent) definitions of "permutation" you are using. $\endgroup$ – Robert Israel Aug 17 '12 at 6:01
  • $\begingroup$ Isomorphic objects in the category of groups. (Or equal objects, depending on your definition of permutation.) $\endgroup$ – Rasmus Aug 17 '12 at 6:01
  • 2
    $\begingroup$ What do you mean by permutations of $n$ things? $\endgroup$ – Qiaochu Yuan Aug 17 '12 at 6:03

Your question is pretty vague, but that's ok since you are merely looking for intuition. Some of the comments answer your question, but I will try to give some more details.

First of all if you talk about an object you want to be very precise what it is, then you should carefully define what you are talking about.

In you example you could define permutation as

1) A permutation is a bijection of a finite ordered set on itself, or

2) A permutation is a sequence containing each element of a finite ordered set once, and only once.

There are probably more definitions but for now these will suffice.

Now note that the first definition here just defines the first object you want to talk about as the second object you mention. So they are virtually the same. If we take the second definition, then you can talk about this object without knowing your second object. But obviously the definitions are equivalent - or two different ways to look at the same object.

To be completely precise we have to specify what we mean by equivalent. In particular we need a world in which we can compare these objects. One way to do this is to understand both objects as groups. Then they are isomorphic as groups which usually as good as being equal for all that matters.

Edit: @celtschk's comment shows that the question is actually deeper. Above I am talking about finite ordered sets. If we work without this extra condition then the two objects aren't even isomorphic as groups. However if we understand these objects as categories, 1) as a category with one object and bijection as morphisms and 2) as a category with one object for each choice of an ordering, and the sequences and additionally a unique isomorphism between any two object as morphisms, then the two categories are equivalent.

Remark: Since I assume that this answer is maybe above the OP's head the moral answer is: Depnding on your definition these objects are either equal, equivalent or not even equivalent. To say what we mean by equivalent we have to fix a universe to work in, which might change whether they are equivalet or not.

  • 1
    $\begingroup$ Actually the two are only equivalent if you also have an order on your finite set. Or else, how do you determine if the sequence $[a,\alpha,\aleph]$ or the sequence $[a,\aleph,\alpha]$ is the identity permutation? $\endgroup$ – celtschk Aug 17 '12 at 10:56
  • $\begingroup$ Thanks @celtschk, this comment revealed that the question is actually even deeper than I initially expected. $\endgroup$ – Simon Markett Aug 17 '12 at 12:03

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