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Let $X$ be a topological space and $\mathcal{C}$ be category of topological spaces over $X$.

  • Objects are topological space with map $\pi:Y\rightarrow X$.
  • Morphisms are continuous maps compatible with the $\pi$ maps.

I am trying to see what are group objects in this category.

To say an object $\pi:G\rightarrow X$ is a group object I need to give multiplication morphism, identity morphism, inverse morphism.

When defining group object, we assume that category has finite products. In this category, product of $\pi_1: G_1\rightarrow X$ and $\pi_2:G_2\rightarrow X$ is $\eta:G_1\times_X G_2\rightarrow X$ where $$G_1\times_X G_2=\{(g_1,g_2):\pi_1(g_1)=\pi_2(g_2)\}$$ and $\eta(g_1,g_2)=\pi_1(g_1)=\pi_2(g_2)$.

In case when $G_1=G_2$, product is just $G\times G$. So, I am guessing group objects are just topological groups with continuous maps to $X$.

Am I missing anything?? This does not seem to be correct.

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    $\begingroup$ No, the product is $G\times_X G$, not $G\times G$. So $G$ is not necessarily a group itself. Take for example the tangent bundle of a manifold $M$. This is a group over $M$ by addition of vectors. But, it is not a group itself : if $(v,x), (w,y)\in TM$ (that is, $x,y\in M$ and $v\in T_xM$, $w\in T_yM$), then what is $(v,x)+(w,y)$ ? On the other hand, there is a well defined map $TM\times_M TM\to TM$ such that $(v,x),(w,x)\mapsto (v+w,x)$. $\endgroup$
    – Roland
    Apr 1, 2018 at 12:20
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    $\begingroup$ It is not true that $G \times_X G = G \times G$. In fact, $G \times_X G = \{ (g_1,g_2) \mid \pi(g_1)=\pi(g_2) \}$. It might help to think of $\pi : G \to X$ as an $X$-indexed family of spaces $(G_x \mid x \in X)$, where $G_x = \pi^{-1}(x)$; then the product is $(G_x \times G_x \mid x \in X)$. $\endgroup$ Apr 1, 2018 at 12:21
  • $\begingroup$ Hehe.. I am mistaken. Though I have written here correctly what $G_1\times_XG_2$ is correctly, I have written in my book something else and that said $G\times_XG=G\times G$ just an overlook. @CliveNewstead thinking of $G\rightarrow X$ as what you said does help to view that better, $\endgroup$
    – user537667
    Apr 1, 2018 at 12:24
  • $\begingroup$ @Roland that is true. It is a silly mistake I have made. Can you say something about group object? $\endgroup$
    – user537667
    Apr 1, 2018 at 12:28
  • $\begingroup$ What do you mean by this question ? Every topological space is a group over itself... $\endgroup$
    – Roland
    Apr 1, 2018 at 12:36

1 Answer 1

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The terminal object of $\mathbf{Top}/X$ is $\mathrm{id}_X : X \to X$, and product is given by pullback. With this in mind, a group object in $\mathbf{Top}/X$ consists of:

  • An object $\pi : G \to X$;
  • A morphism $e : X \to G$ such that $\pi \circ e = \mathrm{id}_X$;
  • A morphism $i : G \to G$ such that $\pi \circ i = \pi$;
  • A morphism $m : G \times_X G \to G$ such that $\pi \circ m = \pi \times_X \pi$;

such that the usual group axioms hold.

As I suggested in the comments above, it helps to think of objects of $\mathbf{Top}/X$ as $X$-indexed families of topological spaces, in which case you can think of $\pi,e,i,m$ as:

  • A family $(G_x \mid x \in X)$ of topological spaces;
  • A family $(e_x \in G_x \mid x \in X)$ of elements of the spaces;
  • A family $(i_x : G_x \to G_x \mid x \in X)$ of continuous maps; and
  • A family $(m_x : G_x \times G_x \to G_x \mid x \in X)$ of continuous maps.

The group axioms can then simply be interpreted componentwise, so that you can draw the conclusion that a group object in $\mathbf{Top}/X$ is an $X$-indexed family of topological groups.

Translating back to the 'usual' setting, a group object in $\mathbf{Top}/X$ is a continuous map $\pi : G \to X$ together with maps $e,i,m$ as above, such that for each $x \in X$, the fibre $\pi^{-1}(x)$ is a topological group under the restrictions of $e,i,m$ to the appropriate domains.

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  • $\begingroup$ This is what I am looking for. Is $\pi:G\rightarrow X$ simply a collection of topological spaces $G_x$ or is there any relation between them? I am asking because there is continuity condition given. If $\pi:G\rightarrow X$ is just a set map then we can see $G$ as collection of topological spaces, in this case of $\pi$ being continuous I think we can say some extra relation. $\endgroup$
    – user537667
    Apr 1, 2018 at 13:51
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    $\begingroup$ @cello: Right, it's really just for the sake of intuition; continuity of the map $\pi : G \to X$ is important, since that's what ties the groups $G_x$ together. $\endgroup$ Apr 1, 2018 at 13:53
  • $\begingroup$ What does it mean to say ties together?? $\endgroup$
    – user537667
    Apr 1, 2018 at 13:56
  • $\begingroup$ @cello: Continuity is what glues together the fibres $G_x = \pi^{-1}(x)$. Saying you have 'an $X$-indexed family $(G_x \mid x \in X)$' tells you nothing about how the groups $G_x$ interact with the topology of $G$. $\endgroup$ Apr 1, 2018 at 14:07
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    $\begingroup$ @Cello: Again, the word 'glues' was just intuition: topological spaces can be pretty whacky... but here's one example: The projection map $\pi : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$, $(a,x) \mapsto x$, gives a group object in $\mathbf{Top}/\mathbb{R}$. Each fibre $\pi^{-1}(x) = \mathbb{R} \times \{ x \}$ is a topological group under addition, i.e. $(a,x) + (b,x) = (a+b,x)$. Without knowing that $\pi$ is continuous, we just have a bunch topological groups $\mathbb{R} \times \{ x \}$ that look like lines; but continuity of $\pi$ (intuitively) 'glues' them together to form a plane. $\endgroup$ Apr 1, 2018 at 14:29

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