16
$\begingroup$

Of the mathematical objects that I am familiar with, it is normally the case that the product of 2 objects is an object of the same type and that an equivalence relation on an object induces a quotient object of the same type.

I think I have some understanding as to why the product of 2 fields is not a field, because a field is not an algebra in the universal algebra sense. But I don't see a reason as to why an equivalence relation on a metric space fails to induce a quotient structure, apart from the fact that it just doesn't work.

Improve this question
$\endgroup$
8
  • 6
    $\begingroup$ Are you thinking about the category $\mathcal{C}$ of metric spaces and continuous maps, or the category $\mathcal{D}$ of metric spaces and metric-preserving maps (ie isometric embeddings)? In $\mathcal{C}$ (which is essentially a full subcategory of topological spaces) I don't think that there is anything interesting to say about your question. Category $\mathcal{D}$ is much more natural if you want to make analogies with algebraic categories. $\endgroup$
    – Neil Strickland
    Commented Mar 6, 2013 at 11:53
  • 3
    $\begingroup$ I'm not sure whether this really answers your question, but the reason why limits and colimits of topological spaces exist is that on a given set there are always the coarsest and the finest topology satisfying some condition. On the other hand it doesn't seem sensible to talk about "coarsest" and "finest" metrics. $\endgroup$
    – Karol Szumiło
    Commented Mar 6, 2013 at 12:47
  • 4
    $\begingroup$ On the other hand, a quotient norm is often seen for normed spaces. $\endgroup$
    – Gerald Edgar
    Commented Mar 6, 2013 at 13:22
  • 7
    $\begingroup$ I'm dubious of the motivation in the first paragraph. Plenty of structures require some conditions for a quotient to be an object of the same type. For example, modding out a group by an equivalence relation, even an equivalence relation arising from a subgroup, does not in general give a group; you need normality. Even in the case of topological spaces, where you can always define quotients (as per Karol's comment), you need conditions for the resulting space to be nice. So your question could just as easily ask about Hausdorff spaces, or many other things, instead of metric spaces. $\endgroup$
    – Noah Stein
    Commented Mar 6, 2013 at 13:44
  • 2
    $\begingroup$ @Noah: I agree in principle, but note that quotients do exist in Hausdorff spaces, as Hausdorff spaces are a reflective subcategory of all topological spaces, hence cocomplete. $\endgroup$
    – David Carchedi
    Commented Mar 6, 2013 at 16:12

2 Answers 2

Reset to default
32
$\begingroup$

You can define a (pseudo)metric on a quotient of a metric space. Let $X$ be a metric space with metric $d$ and an equivalence relation $\sim$. Say that a chain between two points $x,y\in X$ is a sequence of points $x=a_0\sim b_0$, $a_1\sim b_1$, $\ldots$ $a_n\sim b_n=y$, and define the length of such a chain to be $\sum d(b_i,a_{i+1})$. We can now define the distance $d([x],[y])$ between two equivalence classes to be the infimum of all lengths of chains from $x$ to $y$.

It's easy to see that this is a pseudometric on $X/{\sim}$ (a metric where the distance between two distinct points might be $0$). This descends to a true metric on the quotient $Y=X/{\sim}'$, where $x\sim' y$ if $d([x],[y])=0$. Furthermore, $Y$ can be characterized by the following universal property: (non-strictly) distance-decreasing maps from $Y$ to a metric space $Z$ are naturally in bijection with distance-decreasing maps from $f:X\to Z$ such that $f(x)=f(y)$ whenever $x\sim y$.

More generally, a similar construction shows that the category of metric spaces and distance-decreasing maps has all connected colimits (colimits over connected diagrams). If you generalize metrics to allow the distance between two points to be infinite, you can construct all colimits, and also all limits (use the sup metric on products).

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ This kind of limit/quotient is important in the theory of group actions on $\mathbb{R}$-trees, for instance the paper of Levitt and Paulin, "Geometric group actions on trees", MR1428059. $\endgroup$
    – Lee Mosher
    Commented Mar 6, 2013 at 13:24
  • 2
    $\begingroup$ See also ncatlab.org/nlab/show/metric+space#lawvere_metric_spaces_6 . $\endgroup$
    – Qiaochu Yuan
    Commented Mar 7, 2013 at 4:50
  • 3
    $\begingroup$ This construction is discussed in detail in Section 1.4 of my book Lipschitz Algebras. $\endgroup$
    – Nik Weaver
    Commented Dec 31, 2016 at 1:58
  • $\begingroup$ Does this pseudometric simplify to anything in particular when every equivalence class is simply a pair with every pair drawn one element each from a pair of disjoint subsets of $X$ as in this example: math.stackexchange.com/q/3811805/334732 ? $\endgroup$
    – Robert Frost
    Commented Sep 3, 2020 at 14:07
14
$\begingroup$

One way to get around the lack of quotients is to work in the category of Uniform Spaces. This is more general than a metric space, but still gives you a way to talk about how "close" two points are. Uniform spaces are also cool because they simultaneously generalize metric spaces and topological groups. I believe their study was popular back in the 60s and 70s for instance because this category of uniform spaces has much nicer categorical properties than either of the categories it generalizes. For instance, with uniform maps it's complete and cocomplete, as can be seen here.

Even if your maps are just continuous maps, you still get quotients as can be seen here. Isbell's book is a great reference for this subject.

Side note: I like Eric Wofsey's answer a lot and I wonder if it's secretly related to this answer. The wikipedia page points out the connection between uniform spaces and pseudo-metrics but for categorical things it's probably easy to work with the former than the latter.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ Yes, your answer is related to Eric's. The category of metric spaces and uniformly continuous maps has many quotients (though not all) defined by the formula in Eric's answer. (In other words, that formula gives a well-defined metric, in many cases, if the original metric is changed without changing the uniform structure.) For example, you can define mapping cylinder and join in this category (which you can't do for metric spaces and continuous maps). This is done in arxiv.org/abs/1106.3249 $\endgroup$
    – Sergey Melikhov
    Commented Mar 7, 2013 at 3:30

You must log in to answer this question.

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