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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 $f=f^\prime\circ\iota$?

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    $\begingroup$ 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. $\endgroup$
    – goblin GONE
    May 3, 2015 at 10:49
  • $\begingroup$ @goblin Thanks for the idea you provided. How do you think about the original problem? You feel it should hold or not? $\endgroup$
    – Censi LI
    May 3, 2015 at 15:11
  • $\begingroup$ 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 $\mathbf{Top}.$ $\endgroup$
    – goblin GONE
    May 3, 2015 at 16:50
  • $\begingroup$ @goblin I'll read that. Thank you very much. $\endgroup$
    – Censi LI
    May 3, 2015 at 16:53
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    $\begingroup$ When I google "free topological group", the first hit also seems relevant. $\endgroup$
    – tcamps
    May 4, 2015 at 2:35

2 Answers 2

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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.

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  • $\begingroup$ @CensiLI: Any homomorphism is determined by values on a generating set and $i[X]$ is a generating set by the definition. $\endgroup$
    – user87690
    May 3, 2015 at 16:14
  • $\begingroup$ How can you ensure that a set $\mathcal{F}$ exists? Not that I don't trust it, but it's worth mentioning. $\endgroup$
    – egreg
    May 3, 2015 at 16:22
  • $\begingroup$ @CensiLI: Note that the resulting group is not the whole product but just the subgroup generated by $i[X]$. $\endgroup$
    – user87690
    May 3, 2015 at 16:25
  • $\begingroup$ @egreg: It is enough to consider maps that with codomains of bounded size. $\endgroup$
    – user87690
    May 3, 2015 at 16:26
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    $\begingroup$ @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. $\endgroup$
    – Martin Brandenburg
    May 4, 2015 at 15:22
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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.

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  • $\begingroup$ 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. $\endgroup$
    – goblin GONE
    May 5, 2015 at 7:02

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