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

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!

• Which book are you referring to? Apr 29, 2020 at 22:55
• Jantzen - Algebra p.21
– Zest
Apr 29, 2020 at 22:59
• If $\pi$ were not surjective it is possible that one could define a homomorphism which has all the same values on the image of $\pi$ but is not the same on the rest of $G$ Apr 29, 2020 at 23:00
• I dont understand, $\operatorname{Im} \pi \not\subset G$
– Zest
Apr 29, 2020 at 23:03

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.

• beautiful, thank you Arturo.
– Zest
Apr 29, 2020 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, 2020 at 0:03
• @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. Apr 30, 2020 at 0:14
• @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. Apr 30, 2020 at 0:15
• @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 $A$. Apr 30, 2020 at 0:18

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.

• Thanks @ArturoMagidin. I spelled it wrong.
– user403337
Apr 29, 2020 at 23:19