Construct a "natural" bijection between homomorphisms with domain $G$ and homomorphisms with a factor group $G/H$ as domain.

Question

Assume $G$ is a group with a normal subgroup $H$. Let $K$ be another group. Consider the sets $$X = \{ \varphi: G/H \rightarrow K \mid \varphi~\text{ is a homomorphism.} \} $$ and $$Y = \{ \varphi: G \rightarrow K \mid \varphi~\text{ is a homomorphism and takes value}~e_K~\text{on}~H. \}.$$ We want to show that there is a "natural" bijection between the two sets.

To construct a mapping from $X$ to $Y$, we can use the canonical projection $\pi : G \rightarrow G/H$ with $g \mapsto gH$. This yields $X \ni \varphi \mapsto \varphi \circ \pi \in Y$.

How do I construct the inverse of this map?

Context: This is a claim made during a lecture after discussing isomorphism theorems. Constructing the inverse should require something similar to what one does in proving the first isomorphism theorem.

Answer 1

The point is that if you have some $\varphi: G \rightarrow K$ with kernel containing $H$ then there exists some unique $\varphi' : G/H\rightarrow K$ satisfying $$\varphi'(g\cdot H)=\varphi(g)$$ where $g\cdot H$ is a coset in $G/H$. You can check this in the usual way one checks such definitions - first that it is well defined, since if $g\cdot H=g'\cdot H$ then $\varphi(g)=\varphi(g')$, and then that this is a homomorphism (which is not too hard once we know it's well-defined).

This association between functions $\varphi:G\rightarrow K$ with kernel containing $H$ and maps $\varphi':G/H\rightarrow K$ is the map $Y$ to $X$ that you're looking for. You can check that it is inverse to the map you already have.

Answer 2

The first isomorphism theorem tells you that for any group homomorphism $\varphi : G \to K$ with $H \subseteq \ker \varphi$ there exists a unique group homomorphism $\tilde{\varphi} : G/H \to K$ such that $\tilde{\varphi} \circ \pi = \varphi$, where $\pi : G \to G/H$ is the canonical projection, namely, $\tilde{\varphi}(gH) = \varphi(g)$ for every $g \in G$ (check that this is well-defined!). Thus, the map $$X \to Y; \quad\phi \mapsto \phi \circ \pi$$ is a bijection (since for every element of $Y$ there is a unique element of $X$ which is in its preimage).