The universal property of group says that if $G,G'$ are two groups and $f:G\longrightarrow G'$ is a homomorphism, then, there is a unique $\bar f:G/\ker f\longrightarrow G'$ s.t. $f=\bar f\circ \pi$ where $\pi:G\longrightarrow G/\ker f$ the canonical projection. How can I interpret it ? What does it mean exactly ?
What I see is almost that
$$\forall f\in \text{Hom}(G,G'), \exists !\bar f\in \text{Hom}(G/\ker f,G'): f=\bar f\circ \pi.$$
In other word that if $N\trianglelefteq G$ then
$$\text{Hom}(G,G')\cong \text{Hom}(G/N,G'),$$
or something like that, but I don't see exactly what it is neither what it gives.
Thank you for helping (and I hope the question is clear).
Best Answer
In categroy $\mathbb{C}$, Universal object I is object in $\mathbb{C}$ such that for each object A in $\mathbb{C}$, $\exists$! morphism I$\rightarrow $A.
Above thing can be interpreted by universal mapping property by constructing appropriate Category $\mathbb{D}$
Let's construct Category $\mathbb{D}$
for fixed group G, group homomorphism $\textit{f}$ : G $\rightarrow $ G ′
Define object be group homomorphism $\textit{g}$ : G $\rightarrow $ K such that $\textit{g}$(a)=$\textit{g}$(b) $\Leftrightarrow $ a-b$\in $ ker $\textit{f}$
Define morphism : morphism $\textit{g}$ $\rightarrow $$\textit{h}$ (where $\textit{g}$ : G$\rightarrow $K, $\textit{h}$ : G$\rightarrow $H) be group homomorphism $\psi$ : K$\rightarrow $H such that $\psi$$\textit{g}$=$\textit{h}$
then $\mathbb{D}$ is category.(Can be proved!) clearly $\textit{f}$ : G $\rightarrow $ G ′ is object in $\mathbb{D}$
Consider $\pi$ : G⟶G/ker$\textit{f}$. ($\pi$ is object in $\mathbb{D}$ ) Claim for any object $\textit{g}$ : G ⟶ K, ∃!$\phi $ : G/ker$\textit{f}$ ⟶ K
(this imply $\pi$ is universal object in $\mathbb{D}$)
Define $\phi$(a+ker$\textit{f}$)=g(a) then it can be cheked that $\phi$ is well defined unique homomorphism such that $\phi$$\pi$=$\textit{g}$
therefore $\pi$ is universal object in $\mathbb{D}$
Since $\textit{f}$ is object of $\mathbb{D}$, there is unique morphism $\pi$⟶$\textit{f}$ by universal mapping property
$\therefore $ there is unique homomorphism $\overline{f}$ : G/ker$\textit{f}$ ⟶ G ' such that $\overline{f}$$\pi$=$\textit{f}$
"∀$\textit{f}$∈Hom(G, G′),∃!$\overline{f}$ ∈Hom(G/ker$\textit{f}$, G′) such that $\textit{f}$=$\overline{f}$ ∘π." is just categorical language of "for any group homomorphism $\textit{f}$ : G $\rightarrow $ G ′, there is unique homomorphism $\overline{f}$ : G/ker$\textit{f}$⟶ G ' such that $\overline{f}$$\pi$=$\textit{f}$
you can also check $\pi$': G → G/N (where N$\subseteq $ker$\textit{f}$) is universal object in $\mathbb{D}$.
Hom(G,G′)≅Hom(G/N,G′)$\leftrightarrow $there is group isomorphism between Hom(G,G′) and Hom(G/N,G′)
$\eta $ : Hom(G,G′)⟶Hom(G/N,G′) such that $\textit{f}$⟶$\overline{f}$
$\zeta$ : Hom(G/N,G′)⟶Hom(G,G′) such that $\textit{g}$⟶$\textit{g}$$\pi$'
Then you can check that $\eta $ is isomorphism. therefore Hom(G,G′)≅Hom(G/N,G′)