Let $H \trianglelefteq \text{ group } G$ and let $H' \trianglelefteq \text{ group } G'$. Let $\phi$ be a homomorphism of G into G'.
Show that if $\phi[H] \subseteq H'$, then $\phi$ induces a natural homomorphism $♥ : (G/H) \to (G'/H')$.
(This fact is used constantly in algebraic topology.)

Proof: For $gH ∈ G/H$, let $♥(gH) = \phi(g)H$. I shirk the rest.

(1.) I condone the proof hence not asking about it. I'm confounded how you can envisage and envision this homomorphism? Where does it spring from? Feels more magical than natural.

(2.) I don't know any topology but I feel this question's saying this result's important? Hence what's the intuition? Even after proving this, I can't flesh out what this exercise and proof mean.

(3.) Can someone please draw a picture of all this? Thanks a lot.


(1) It is natural because it is virtually unavoidable to find it if you are looking for a homomorphism $G/H\to G'/H'$: Such a homomorphism takes an element of $G/H$ as input. These elements are of the form $gH$. The only thing that identifies such an element is $g$, so proceed with that. You must somehow find an element of $G'/H'$, i.e. some $g'H'$ with $g'\in G'$. In essence, we need an element of $G'$, given an element $g\in G$. The only way to get from $G$ to $G'$ in the present situation is by iusing $\phi$. So the natural choice is to pick $g'=\phi(g)$. In total the natural way to obtain an element of $G'/H'$ from an element of $G/H$ is to map $gH\mapsto \phi(g)H'$. Then verify that this is well-defined: This requires precisely that for $g\in H$ we have $\phi(g)\in H'$, yay!

(2) This result (and others with homomorphisms aoccuring naturally) are important everywhere, not just algebraic topology.

(3) Maybe the best picture possible would require you to diagram-hunt, which I am afraid you have not covered yet: $$\begin{matrix}0&\longrightarrow&H&\longrightarrow&G&\longrightarrow&G/H&\longrightarrow &0\\ &&\downarrow&&\downarrow&&\downarrow\\ 0&\longrightarrow&H'&\longrightarrow&G'&\longrightarrow&G'/H'&\longrightarrow &0 \end{matrix} $$


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