Dual statement to the fundamental theorem of group homomorphism? Let $G$ and $H$ be groups. The fundamental theorem of group homomophism states that, for any surjection $f : G \to H$, there exists an isomorphism $\eta : G/\text{ker}(f) \to H$.
I am wondering if there exists a dual statement to this theorem: starting with an injection $f : G \to H$. It would be interesting to see $\text{im}(f)$ (or the cokernel of some map) appearing instead of $\text{ker}(f)$, in some way.
I've read that surjections and injections exist in opposite categories; I've thus tagged 'category theory', for potential reasoning in this area.
 A: You can dualize the fundamental theorem on homomorphisms in its more general form. It states that for a surjective homomorphism of groups $p : G \twoheadrightarrow Q$, a homomorphism $f : G \to H$ factors through $p$ (i.e., we have $f = \overline{f} \circ p$ for some homomorphism $\overline{f} : Q \to H$) if and only if $\ker(p) \subseteq \ker(f)$. This applies, in particular, to the special case of the projection $p : G \twoheadrightarrow G/N$ with $\ker(p) = N$, in which we get the usual statement that $f : G \to H$ factors through $G \twoheadrightarrow G/N$ iff $N \subseteq \ker(f)$.
The dualization is as follows: If $i : U \hookrightarrow G$ is an injective homomorphism of groups, then a homomorphism of groups $f : H \to G$ factors through $i$ (i.e., $f = i \circ \overline{f}$ for some homomorphism $\overline{f} : H \to U$) if and only if $\mathrm{im}(f) \subseteq \mathrm{im}(i)$. This applies, in particular, to the special case of an inclusion map of a subgroup $U \subseteq G$. So here, $f : H \to G$ factors through $U \hookrightarrow G$ if and only if $\mathrm{im}(f) \subseteq U$.
This dual statement is very easy to prove and often tacitly used, but usually not stated explicitly in textbooks on group theory or algebra, probably because it is so obvious.
These statements are, in fact, dual to each other, since the first one is a criterion for factoring homomorphisms through epimorphisms of groups, whereas the second one is about monomorphisms of groups.
Similar statements hold for all algebraic structures (sets, rings, abelian groups, lattices, Lie algebras, etc.). Here, the kernel has to be replaced by the kernel congruence relation $\ker(f) := \{(x,x') : f(x)=f(x')\}$, but the image stays the same.
But the statements do not hold verbatim for, say, topological spaces. The first statement holds in $\mathbf{Top}$ iff $Q$ carries the final topology with respect to $q$, and the second statement holds iff $U$ carries the initial topology with respect to $i$ (because we need $\overline{f}$ to be continuous in each case).
