Why does the surjectivity of the canonical projection $\pi:G\to G/N$ imply uniqueness of $\tilde \varphi: G/N \to H$

Question

Let's look at the universal property of quotient groups:

Let $\varphi:G \to H$ be a homomorphism, $N$ a normal subgroup of $G$ and $\pi:G \to G/N$ the canonical projection. If $N \le \ker \varphi$, there is a homomorphism $\tilde \varphi: G/N \to H$ such that $$\tilde \varphi \circ \pi = \varphi.$$

The proof in my book for this universal property states that the uniqueness of $\tilde \varphi$ follows directly from the surjectivity of the canonical projection $\pi$.

However, i don't understand, how and why.

How would a comprehensive proof of the uniqueness of $\tilde \varphi$ look like, rather than stating "follows from surjectivity"? In other words:

Why would $\tilde \varphi$ not be unique if $\pi$ wasn't surjective?

Thanks for any help!

Answer 1

Surjective functions can be cancelled on the right:

If $f\colon A\to B$ is surjective, and $g,h\colon B\to C$ are functions such that $g\circ f = h\circ f$, then $g=h$. For, given $b\in B$, there exists $a\in A$ such that $f(a)=b$. Therefore, $$g(b) = g(f(a)) = h(f(a)) = h(b),$$ so $g(b)=h(b)$ for all $b\in B$, hence $g=h$.

(In fact, this property characterizes surjective functions in set theory).

So suppose that you have two functions, $\overline{\varphi},\overline{\phi}\colon G/N\to H$ such that $\overline{\varphi}\circ \pi = \overline{\phi}\circ \pi$. Since $\pi$ is surjective, this immediately implies that $\overline{\varphi}=\overline{\phi}$.

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“How would $\overline{\phi}$ not be unique if $\pi$ wasn’t surjective?” is a counterfactual question.

But...

In general, if $H$ is a proper subgroup of $G$, then there always exists a group $K$ and group homomorphisms $f,g\colon G\to K$ such that $f(h)=g(h)$ for every $h\in H$, but $f\neq g$. A construction is given here.

So if, somehow (complete counterfactual, but whatever), $\pi \colon G\to G/N$ were not surjective, then you would be able to construct a group $H$ and homomorphism $f,g\colon G/N \to H$ such that $f\circ\pi = g\circ\pi$, but $f\neq g$. Letting $\varphi=f\circ \pi$ would give you a map with $N\subseteq \mathrm{ker}(\varphi)$, but with both $f,g\colon G/N\to K$ satisfying the conclusion.

Answer 2

You've got a commutative diagram. There's only one way that works, letting $\tilde\phi(gN)=\phi(π^{-1}(gN)$. $π$ surjective means $π^{-1}(gN)\ne\emptyset$. And $N\le\operatorname{ker}\phi$ ensures it is well defined.