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!

  • $\begingroup$ Which book are you referring to? $\endgroup$
    – Shaun
    Commented Apr 29, 2020 at 22:55
  • $\begingroup$ Jantzen - Algebra p.21 $\endgroup$
    – Zest
    Commented Apr 29, 2020 at 22:59
  • $\begingroup$ 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 $G$ $\endgroup$
    – Sempliner
    Commented Apr 29, 2020 at 23:00
  • $\begingroup$ I dont understand, $\operatorname{Im} \pi \not\subset G$ $\endgroup$
    – Zest
    Commented Apr 29, 2020 at 23:03

2 Answers 2


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.


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.

  • $\begingroup$ beautiful, thank you Arturo. $\endgroup$
    – Zest
    Commented Apr 29, 2020 at 23:07
  • $\begingroup$ 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 $g \not=h$? $\endgroup$
    – Zest
    Commented Apr 30, 2020 at 0:03
  • $\begingroup$ @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. $\endgroup$ Commented Apr 30, 2020 at 0:14
  • $\begingroup$ @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. $\endgroup$ Commented Apr 30, 2020 at 0:15
  • $\begingroup$ @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 $A$. $\endgroup$ Commented 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.

  • $\begingroup$ Thanks @ArturoMagidin. I spelled it wrong. $\endgroup$
    – user403337
    Commented Apr 29, 2020 at 23:19

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