Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm learning some category theory and I thought I had understood universal objects but maybe I have not because I cannot write down the definition of initial topology in terms of categories.

My understanding of universal objects so far: I thought universal objects are either initial or terminal objects in a comma category $(O \downarrow F)$ where $O$ is a selection functor $\textbf{1} \to O \in C$ and $F$ is a forgetful functor into $C$. In the example of a free group, $F: \textbf{Group} \to \textbf{Set}$ and $O$ is the set that generates the free group. The objects in this category are pairs $(f, G)$ where $f$ is a (set-theoretic) map from $S \to G$.

Now I would like to do the same for the initial topology but I'm struggling to see what the objects are and also what $F$ should be. $O$ is probably going to be $X$, the space we would endow with the topology. So we probably want $F$ to be a functor $\textbf{Top} \to \textbf{Set}$. The objects are probably $(f,Y)$ where $f$ is a map $X \to Y$ (or $Y \to X$?) such that there exists a continuous map $f^\prime$ with $F(f^\prime) = f$.

Is this right? If not: what are the objects? And is the initial topology an initial or a terminal object?


I think what I wrote above is wrong: I get one commutative diagram for each $Y_i$ in the family with respect to which the initial topology is defined. This means that the objects are $(g_i, Z)$ where $g_i : Z \to Y_i$ and the comma category would be $(Y_i, F)$. Is this right?

The diagram for $(Y_i \downarrow F)$ would look like this:

X --(f_i)--> Y_i

Then the question is whether the unique continuous map $\varphi$ is $Z \to X$ or $X \to Z$.

share|cite|improve this question
If I'm not mistaken,, is what you want. – Martin Wanvik Apr 29 '12 at 9:07
@MartinWanvik Thank you! It might be but at the moment I don't understand yet whether this is the same as what I wrote above or not because I don't know much about category theory yet and I don't know what cones are or adjoint functors. I will have to think about this. – Rudy the Reindeer Apr 29 '12 at 9:48
I'll write an answer, trying to expand a bit on the description given on the wikipedia page. – Martin Wanvik Apr 29 '12 at 9:57
@ClarkKent: The universal property of the initial topology looks like a limit, so if it's going to be given by a limit, it has to be a terminal object! – Zhen Lin Apr 29 '12 at 15:16
@ZhenLin Thank you very much for pointing this out! – Rudy the Reindeer May 3 '12 at 10:41
up vote 9 down vote accepted

(This is an expanded version of this).

First of all, the given collection of spaces $\{ Y_j \}_{j \in J}$ can be viewed as a functor $Y: \mathbf{J} \to \mathbf{Top}$, where $\mathbf{J}$ is simply the indexing set $J$ viewed as a discrete category (i.e. one whose objects are $J$ and containing only identity morphisms). In other words, we have $Y(j) = Y_j$ and $Y(\operatorname{id}_j) = \operatorname{id}_{Y(j)}$, which is trivially a functor. The other part of the description of the initial topology is the collection of maps $f_j: X \to Y_j$ where $X$ is a set. More formally, if $U: \mathbf{Top} \to \mathbf{Set}$ is the forgetful functor, then we can also write $f_j: X \to U(Y_j)$ (this takes place in the category $\mathbf{Set}$). As the article mentions, the collection $\{f_j\}$ can be viewed as a cone from $X$ to $UY$. Here comes the definition:

Given categories $\mathbf{C},\mathbf{D}$, and a functor $F: \mathbf{D} \to \mathbf{C}$, a cone from an object $X$ in $\mathbf{C}$ is a natural transformation $\eta: \Delta(X) \Rightarrow F$, where $\Delta: \mathbf{C} \to \mathbf{C^D}$ is the functor mapping each object $Z$ in $\mathbf{C}$ to the constant functor $\Delta(X): \mathbf{D} \to \mathbf{C}$ (which takes each object in $\mathbf{D}$ to $Z$ and each morphism in $\mathbf{D}$ to the identity on $Z$). If $f: V \to W$ in $\mathbf{C}$, then $\Delta(f)$ is the natural transformation $\Delta(f): \Delta(V) \Rightarrow \Delta(W)$ whose component at each object is $f$. For each functor $F$, we define the category of cones $\mathbf{Cone}(F)$ as the comma category $(\Delta \downarrow F)$.

Now, it is easy to see that the collection $f_j: X \to UY(j)$ can be viewed as a cone from $X$ to $UY$, i.e an object in $\mathbf{Cone}(UY)$. In other words, it defines a natural transformation $f: \Delta(X) \Rightarrow UY$ (here $\Delta(X)$ is a functor on the category $\mathbf{J}$, described previously). Similarly, a collection of continuous maps $g_j: Z \to Y_j$ defines an object $\mathbf{Cone}(Y)$. Applying the forgetful functor $U: \mathbf{Top} \to \mathbf{Set}$, we get a forgetful functor $U': \mathbf{Cone}(Y) \to \mathbf{Cone}(UY)$.

Now we are in a position to describe the universal property of the initial topology: as mentioned, we start with an object $f_j: X \to UY$ in $\mathbf{Cone}(UY)$. We will write this as $(X,f)$, where $$f = \{f_j: X \to Y_j\}.$$ Now, a morphism of cones $\tau: (Z,g) \Rightarrow (X,f)$ is simply a morphism $\tau: Z \to X$ such that $f_j \circ \tau = g_j$. Imposing the initial topology (coming from the maps $f_j$) on $X$, we get an object $I(X,f)$ in $\mathbf{Cone}(Y)$. Moreover, the identity map defines a morphism $\varepsilon: U'(I(X,f)) \Rightarrow (X,f)$ (in $\mathbf{Cone}(UY)$), which is universal in the following sense: whenever $\eta: U'(Z,g) \Rightarrow (X,f)$ is another morphism in $\mathbf{Cone}(UY)$, there exists a unique morphism $\xi: (Z,g) \Rightarrow (X,f)$ such that $\eta = \varepsilon \circ \xi$. In other words, if we are given

a topological space $Z$, a collection of continuous maps $g_j: Z \to Y_j$, and a map (a morphism in $\mathbf{Set}$) $\eta: Z \to X$ such that $f_j \circ \eta = g_j$,


there exists a (unique) continuous map $\xi: Z \to X$, satisfying $f_j \circ \xi = g_j$ for all $j \in J$, such that $\varepsilon \circ \xi = \eta$.

Of course, since $\varepsilon$ is secretly the identity map on $X$, this implies that $\xi = \eta$ (as set functions, or more accurately, $U(\xi) = \eta$), so all this is saying is that $\xi$ is continuous if $f_j \circ \eta = f_j \circ \xi$ is continuous for all $j \in J$ ($f_j \circ \xi = g_j$, which is the original collection of continuous maps specified above). The converse is of course immediate, since a composite of continuous maps is again continuous. So this highly elaborate categorical language is only saying that

a map $\eta: Z \to X$ is continuous if and only if $f_j \circ \eta$ is continuous for all $j \in J$.

This also implies that imposing the initial topology defines a functor $I: \mathbf{Cone}(UY) \to \mathbf{Cone}(Y)$, and this is right adjoint to the functor $U'$ (and, in fact any right adjoint functor can be described by a similar universal property), but that is another story.

share|cite|improve this answer
Thank you very much. I will take some time to read it! – Rudy the Reindeer May 3 '12 at 10:41
So, is the initial topology an initial object or not?? – Alexander Frei Mar 22 '14 at 17:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.