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 $X \to Z$.

  • $\begingroup$ If I'm not mistaken, en.wikipedia.org/wiki/Initial_topology#Categorical_description, is what you want. $\endgroup$ Apr 29, 2012 at 9:07
  • $\begingroup$ @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. $\endgroup$ Apr 29, 2012 at 9:48
  • $\begingroup$ I'll write an answer, trying to expand a bit on the description given on the wikipedia page. $\endgroup$ Apr 29, 2012 at 9:57
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    $\begingroup$ @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! $\endgroup$
    – Zhen Lin
    Apr 29, 2012 at 15:16
  • $\begingroup$ @ZhenLin Thank you very much for pointing this out! $\endgroup$ May 3, 2012 at 10:41

1 Answer 1


(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.

  • $\begingroup$ Thank you very much. I will take some time to read it! $\endgroup$ May 3, 2012 at 10:41
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    $\begingroup$ So, is the initial topology an initial object or not?? $\endgroup$ Mar 22, 2014 at 17:09
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    $\begingroup$ @AlexanderFrei $\big( I(X,f), \epsilon \big)$ is a terminal "arrow", i.e. a terminal object in the comma category $(U' \downarrow X)$. That's funny actually... (initial in the opposed category I guess) $\endgroup$
    – Noix07
    Jul 6, 2017 at 22:38
  • $\begingroup$ Just to let you know I used your beautiful answer to expand the wiki article on the initial topology. Thanks a lot. :) $\endgroup$ May 6, 2019 at 3:47
  • $\begingroup$ @Noix07 I'm reading Riehl's book and she says: The most basic formulation of a universal property is to say that a particular object defines an initial or terminal object in its ambient category, is she talking about being an initial or terminal object in the comma category? $\endgroup$
    – John Mars
    Apr 9, 2021 at 2:34

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