19

Let TopGrp be the category of topological groups (not necessarily $T_0$) and Top the category of topological spaces. Does the forgetful functor $U:\mathbf{TopGrp}\to\mathbf{Top}$ admit a left adjoint?

To be concrete, given an arbitrary topological space $X$, my question is that, can we find a topological group $\Gamma X$ and a continuous map $\iota:X\to\Gamma X$ such that, for any topological group $G$ and any continuous map $f:X\to G$, there exists a unique continuous group homomorphism $f^\prime:\Gamma X\to G$ satisfying $f=f^\prime\circ\iota$?

Najib Idrissi
  • 56,269
Censi LI
  • 6,075
  • 2
    I've wondered about this for so long. A good answer would be of the form: Theorem. Let $T$ denote a Lawvere theory. Let $\mathbf{C}$ denote a finite-product category enjoying the following properties... (a),(b),(c)... Then the forgetful functor from the category of $T$-models in $\mathbf{C}$ to $\mathbf{C}$ has a left-adjoint. – goblin GONE May 03 '15 at 10:49
  • @goblin Thanks for the idea you provided. How do you think about the original problem? You feel it should hold or not? – Censi LI May 03 '15 at 15:11
  • Yep, I think the left-adjoint should exist. You should have a read about Lawvere theories (google it). Note that $\mathbf{TopGrp}$ is the category of models of the Lawvere theory $\mathsf{Grp}$ of groups in the finite-product category $\mathbf{Top}.$ – goblin GONE May 03 '15 at 16:50
  • @goblin I'll read that. Thank you very much. – Censi LI May 03 '15 at 16:53
  • 1
    When I google "free topological group", the first hit also seems relevant. – tcamps May 04 '15 at 02:35

2 Answers2

9

Let $\mathcal{F}$ be a set of continuous maps from $X$ such that $cod(f)$ is a topological group for each $f ∈ \mathcal{F}$ and every continuous map from $X$ to a topological group factors thought some map from $\mathcal{F}$. Such set exists since it is enough to consider maps whose images generate the codomain group, so their cardinalities are bounded. Consider $i: X \to ∏_{f ∈ \mathcal{F}} cod(f)$ defined as $i(x)(f) = f(x)$.

Clearly, $∏_{f ∈ \mathcal{F}} cod(f)$ is a topological group and $π_f \circ i$ is continuous for each $f ∈ \mathcal{F}$, so $i$ is continuous.

Every continuous $g: X \to G$ factors through some $f ∈ \mathcal{F}$, and for $g'$ defined as $g'(α) = π_f(α) = α(f): X \to cod(f) ⊆ G$ we have $g'(i(x)) = i(x)(f) = f(x) = g(x)$. Clearly, $g'$ is continuous homorphism.

It is enough to consider the subgroup of $∏_{f ∈ \mathcal{F}} cod(f)$ generated by $i[X]$, on which we have also the uniqueness of the extension.

We may also observe, that $i$ is always injective and it is a topological embedding if $X$ is completely regular.

Adam Bartoš
  • 9,403
  • @CensiLI: Any homomorphism is determined by values on a generating set and $i[X]$ is a generating set by the definition. – Adam Bartoš May 03 '15 at 16:14
  • How can you ensure that a set $\mathcal{F}$ exists? Not that I don't trust it, but it's worth mentioning. – egreg May 03 '15 at 16:22
  • @CensiLI: Note that the resulting group is not the whole product but just the subgroup generated by $i[X]$. – Adam Bartoš May 03 '15 at 16:25
  • @egreg: It is enough to consider maps that with codomains of bounded size. – Adam Bartoš May 03 '15 at 16:26
  • @user87690 Yes, if $f\colon X\to G$ is a map into a group, then the subgroup generated by the image of $f$ has cardinality at most $|X|$. But, as I said, it's worth mentioning it. – egreg May 03 '15 at 16:29
  • @CensiLI: That doesn't matter. The resulting map goes from the subgroup generated by $i[X]$ and such map is uniquely determined by the values on $i[X]$, which are prescribed by $g$. – Adam Bartoš May 03 '15 at 16:29
  • @egreg: You are right, I'll add it. – Adam Bartoš May 03 '15 at 16:30
  • @CensiLI: There is unique homomorphism $g'$ which satisfies $g' \circ i = g$, because the image of $i$ generates $Γ$. That is a general fact from algebra. – Adam Bartoš May 03 '15 at 16:43
  • @user87690 I'm sorry but I've found that I didn't really understand your construction of $g'$ at last... Since $∏{f ∈ \mathcal{F}} cod(f)$ is a product but not a coproduct, how can you define a map from $∏{f ∈ \mathcal{F}} cod(f)$ to $G$ by simply designate it's value at some particular point? – Censi LI May 03 '15 at 17:02
  • @CensiLI: For each $α ∈ ∏_{f ∈ \mathcal{F}} cod(f)$ we define $g'(α) := α(f_0)$. $f_0$ is a chosen map through which $g$ factorizes. What is the problem? – Adam Bartoš May 03 '15 at 17:29
  • 1
    @CensiLI: Perhaps you should first understand the abstract construction of the free group on a set. This is also more well-known and explained in the literature. For this, replace "topological group" by "group" in the answer by user87690. – Martin Brandenburg May 04 '15 at 15:22
  • @user87690 Thanks for all your elaboration. – Censi LI May 05 '15 at 11:46
4

If $\mathcal{C}$ is a cocomplete cartesian closed category, then the forgetful functor $\mathsf{Grp}(\mathcal{C}) \to \mathcal{C}$ has a left adjoint. More generally, if $\tau$ is any type of algebraic structures, the forgetful functor $\mathsf{Alg}_{\tau}(\mathcal{C}) \to \mathcal{C}$ has a left adjoint. This is also what goblin mentioned in the comments. (This result should be well-known, but right now I have no reference for it.)

Unfortunately, $\mathsf{Top}$ is not cartesian closed. But the category of compactly generated weak Hausdorff spaces $\mathsf{CGWH}$ is cartesian closed and is (in contrast to $\mathsf{Top}$) a convenient category of topological spaces. It follows that $\mathsf{Grp}(\mathsf{CGWH}) \to \mathsf{CGWH}$ has a left adjoint. Also, $\mathsf{Grp}(\mathsf{CGWH})$ is complete and cocomplete.

Martin Brandenburg
  • 181,922
  • One idea would be to choose a category bigger than $\mathbf{Top}$ such as $\mathbf{Conv}$ that is both cocomplete and cartesian closed. We use the theorem stated in your opening sentence to conclude that the forgetful functor $U:\mathbf{ConvGrp} \rightarrow \mathbf{Conv}$ has a left-adjoint $F$, and then try to argue that if $X$ is an object of $\mathbf{Conv}$ that just happens to be a topological space, then the $UFX$ also happens to be a topological space. I think this would allow us to conclude that the forgetful functor $\mathbf{TopGrp} \rightarrow \mathbf{Top}$ has a left adjoint. – goblin GONE May 05 '15 at 07:02