A morphism $f: X \to Y$ is an epimorphism if for all $g, h: Y \to Z$, if $g \circ f = h \circ f$ then $g = h$. The epimorphisms in the category of groups are the surjective group homomorphisms. The epimorphisms in the category of topological spaces are the continuous surjections. Are the epimorphisms in the category of topological groups the continuous surjective homomorphisms?
What are the epimorphisms in the category of topological groups
category-theoryepimorphismstopological-groups
Related Solutions
It is obvious that injective (respectively, surjective) group homomorphisms are monomorphisms (respectively, epimorphisms).
Monomorphisms in the category of groups are injective maps. Indeed, suppose $\phi\colon G\to H$ is a monomorphism; consider $\alpha\colon\ker\phi\to G$, the canonical injection, and $\beta\colon\ker\phi\to G$, $\beta(x)=1$. Then $\phi\circ\alpha=\phi\circ\beta$: what does $\alpha=\beta$ entail?
Epimorphisms in the category of groups are surjective, but this is a bit more difficult to show (one needs to define an action on the set of cosets of $H$ by the image of $\phi$).
The standard example of a nonsurjective epimorphism in a category is the embedding $\mathbb{Z}\to\mathbb{Q}$ in the category of rings, which is both a monomorphism (obvious) and an epimorphism (try it).
I once read somewhere that the epimorphisms in the category of Hausdorff spaces are precisely those continuous maps with dense image. So as a conjecture I began with that and here’s a solution (it turns out the epimorphisms are precisely the continuous maps with dense image by the way): Suppose $f$ is epi but $f$ does not have dense image. Then there is some $y \in Y$ such that $y \notin \overline{f(X)}$. Urysohn’s lemma now gives that there exists a continuous map $g : Y \mapsto [0,1]$ such that $g(y)=0$ and $g(z)=1$ for all $z \in \overline{f(X)}$. In particular $g(f(x))=1$ for all $x \in X$. So $$g \circ f =1 \circ f$$ where $1$ is the constant $1$ function. So $g=1$, so $0=g(y)=1$, contradiction. We conclude the image of $f$ lies dense in $Y$.
Suppose now $f$ has dense image. Now if $g_1,g_2: Y \rightarrow Z$ are such that they are equal on $f(X)$, $g_1$ and $g_2$ are two continuous maps that are equal on a dense subset, so if we’re lucky they are equal ;). So if we prove the following we’re done:
Lemma Let $h_1,h_2: A \rightarrow B$ be continuous maps between Hausdorff topological spaces that are equal on a dense subset. Then $h_1=h_2$.
Proof We will show the set $C=\{ x \in A | h_1(x) \neq h_2(x) \}$ is open in $A$, from this the result immediately follows. Since $A$ is Hausdorff, the diagonal $\Delta \subset A \times A$ is closed. So the inverse image of $\{(x,y) | x \neq y\}$ under the composition $$A \xrightarrow{x \mapsto (x,x)} A \times A \xrightarrow{(h_1,h_2)} A \times A$$ is open. But this inverse image is precisely $C$ and we’re done.
Best Answer
A cool answer is this one. Sorry if this the same proof given in the reference in the commentaries.
Consider the forgetful functor
$$U: TopGrp \rightarrow Grp$$
For any group $G$ let $L(G)$ be the the same group with the discrete topology and $R(G)$ be with the codiscrete topology. Then
$$TopGrp(L(G),H)=Grp(G,U(H))$$
since every function $L(G)\rightarrow H$ is continuous. We also have that
$$TopGrp(G,R(H))=Grp(U(G),H)$$
since every function $G \rightarrow R(H)$ is continuous.
Thus we conclude that $U$ is both right and left adjoint. Right adjoints preserve monomorphisms and left adjpints preserve epimorphisms. We conclude that a morphism of topological groups is an (mono)epimorphism if, and only if, its underlying morphism of groups is.