Does the forgetful functor from the category of topological rings to the category of topological abelian groups, $U: \mathbf{TopRing} \to \mathbf{TopAb}$, have a left adjoint and if so, what is it? A reference would be very much appreciated.
Left adjoint to forgetful functor from topological rings to topological abelian groups
adjoint-functorscategory-theoryreference-requesttopological-groupstopological-rings
Related Solutions
Whenever I say graph in this answer, I mean simple graph.
We need to be a bit more precise here, and specify what the arrows are in $\mathbf{Gph}$. A natural choice would be maps $f: G \to G'$ between the vertex sets such that if there is an edge between $x,y \in G$ (which I will denote by $E(x, y)$), then there is an edge between $f(x)$ and $f(y)$. In this case the forgetful functor $U: \mathbf{Gph} \to \mathbf{Set}$, which sends a graph to its underlying vertex set, does indeed have a left adjoint.
The construction was already mentioned by Malice Vidrine in the comments. We can define $F: \mathbf{Set} \to \mathbf{Gph}$ by sending a set $X$ to the graph $F(X)$ with vertex set $X$ and no edges. A function $f: X \to Y$ of sets is then also an arrow $f: F(X) \to F(Y)$ in $\mathbf{Gph}$, so we just set $F(f) = f$.
Let $X$ be a set and $G$ be a graph. Then a function $X \to U(G)$ is literally the same thing as a morphism of graphs $F(X) \to G$. So $\operatorname{Hom}(X, U(G)) = \operatorname{Hom}(F(X), G)$, which is definitely natural, so $F$ is left adjoint to $G$.
In fact, $F: \mathbf{Set} \to \mathbf{Gph}$ itself has again a left adjoint $C: \mathbf{Gph} \to \mathbf{Set}$. Here $C$ is the connected components functor. So it takes a graph $G$ to the set of connected components of $G$. It is a good exercise to define $C$ on the arrows in $\mathbf{Gph}$ and to check that it is indeed left adjoint to $F$.
Limits of (commutative) monoids can simply by constructed with pointwise operations from the underlying limits of sets. This shows immediately that $\mathbf{CMon}$ is complete and that $U$ is continuous. To show the solution set condition, let $M$ be a commutative monoid and let $f : S \to U(M)$ be a map. Consider the submonoid $M'$ of $M$ generated by the image of $f$. Explicitly, $M'$ consists of all $\mathbb{N}$-linear combinations of elements of the form $f(s)$, where $s \in S$. The next step is to bound the size of the underlying set of $M'$. This is usually done with cardinal numbers, but actually one does not need cardinal numbers at all for this argument. Notice that there is a surjective map*
$\coprod_{n \in \mathbb{N}} (\mathbb{N} \times S)^n \to U(M').$
which maps $((m_1,s_1)),\dotsc,(m_n,s_n))$ to $m_1 f(s_1) + \cdots + m_n f(s_n)$. It follows that $U(M')$ is isomorphic to a subset of the set $T := \coprod_{n \in \mathbb{N}} (\mathbb{N} \times S)^n$ (which only depends on $S$). By structure transport, it follows that $M'$ is isomorphic to a monoid whose underlying set is a subset of $T$. But there is only a set of such monoids. Therefore, a solution set consists of those monoids whose underlying set is a subset of $T$.
By the way, a similar argument shows that for every finitary algebraic category $\mathcal{C}$ the forgetful functor $\mathcal{C} \to \mathbf{Set}$ has a left adjoint. In fact, it is even monadic.
For commutative monoids, though, the simplest way is just a direction construction: the underlying set of the free commutative monoid on $S$ is the set of functions $S \to \mathbb{N}$ with finite support (these formalize linear combinations with $\mathbb{N}$-coefficients in $S$), and the operation comes from $\mathbb{N}$. For monoids and groups (or lie algebras etc.), direct constructions are not so straight forward anymore and the above method gives a really slick existence proof.
*Alternatively, you can use a surjective map $\coprod_{n \in \mathbb{N}} S^n \to U(M')$.
Best Answer
Yes.
If you have a functor from a topological category, and that functor has a left adjoint if you forget about the topological structure, then frequently (under quite mild assumptions) you can "lift" that left adjoint to the topological case.
One nice reference is Tholen's On Wyler Taut's Lift Theorem, which proves an even more general version of this result. Here's the special case of interest (originally due to Wyler):
Given a commutative diagram of functors
where $T$ and $T'$ are topological and $\tilde{U}$ preserves initial sources, then if $U$ has a left adjoint, $\tilde{U}$ does too!
For us, $A$ and $A'$ are the categories of topological rings and groups (respesctively). These are both topological over rings, and groups, which are our $X$ and $X'$.
Then $U$ and $\tilde{U}$ should be the forgetful functors, and it's clear that initial topologies get sent to initial topologies by $\tilde{U}$ (since it's not touching the topology at all!).
So the left adjoint to $U$ lifts to a left adjoint of $\tilde{U}$, as desired.
I hope this helps ^_^