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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!

Zest
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2 Answers2

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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 $\overline{\varphi}=\overline{\phi}$.

“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.

Arturo Magidin
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  • beautiful, thank you Arturo. – Zest Apr 29 '20 at 23:07
  • do you happen to know an example for a function $f:A \to B$ that is not surjective with two functions $g,h: B\to C$ such that $f\circ g = f \circ h$ but due to $f$ not being surjective $g \not=h$? – Zest Apr 30 '20 at 00:03
  • @Zest: As I said, that property characterizes surjective functions. If $f\colon A\to B$ is not surjective, let $C={0,1}$. Define $g\colon B\to C$ by $g(b)=0$ for all $b$; define $h\colon B\to C$ by $g(b)=0$ if $b\in f(A)$, and $g(b)=1$ otherwise. – Arturo Magidin Apr 30 '20 at 00:14
  • @Zest: For groups it’s more complicated (because the functions have to be morphisms), but I linked to a construction. You can find it there. – Arturo Magidin Apr 30 '20 at 00:15
  • @Zest: And you have the composition going the wrong way. It’s $g\circ f = h\circ f$. Given your definitions, $f\circ g$ is nonsense: the range of $g$ is $C$, but the domain of $f$ is $A$. – Arturo Magidin Apr 30 '20 at 00:18
  • @Zest: For set theoretic functions, look at this and maybe try to prove the results yourself? At least the ones that don’t invoke proving an equivalence with the Axiom of Choice? They are both rather basic and extremely important. – Arturo Magidin Apr 30 '20 at 00:20
  • Thanks for your help Arturo, sorry that was a typo regarding the composition. – Zest Apr 30 '20 at 00:22
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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.

Arturo Magidin
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