[proofplan] The candidate map $\bar{\varphi}: G/N \to H$ is forced: the factorisation $\varphi = \bar{\varphi} \circ \pi$ means $\bar{\varphi}(gN) = \varphi(g)$, and this same equation simultaneously gives uniqueness. The work is to check that the assignment $gN \mapsto \varphi(g)$ does not depend on the choice of representative $g$ — this is where the hypothesis $N \subset \ker \varphi$ is consumed. Once well-definedness is in hand, the homomorphism property of $\bar{\varphi}$ transfers from $\varphi$ via the coset multiplication rule on $G/N$ ([Quotient Group](/theorems/790)), and the factorisation identity is a direct evaluation. The injectivity and surjectivity characterisations are then short coset computations using $\ker\bar\varphi$ and $\operatorname{im}\bar\varphi$. [/proofplan] [step:Define the candidate map and verify it is independent of coset representative] Define \begin{align*} \bar{\varphi}: G/N &\to H \\ gN &\mapsto \varphi(g). \end{align*} We must check this rule does not depend on the representative chosen for the coset. Suppose $gN = g'N$. Then $g^{-1}g' \in N$ (the standard characterisation of equal left cosets), and since $N \subset \ker\varphi$ by hypothesis, $\varphi(g^{-1}g') = e_H$, where $e_H$ is the identity of $H$. Using that $\varphi$ is a homomorphism, \begin{align*} e_H &= \varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g'), \end{align*} so $\varphi(g) = \varphi(g')$. Hence $\bar\varphi(gN)$ depends only on the coset, not on the representative, and $\bar\varphi$ is a well-defined function. [guided] The defining equation $\varphi = \bar\varphi \circ \pi$ forces the value of $\bar\varphi$ on the coset $gN = \pi(g)$ to be $\varphi(g)$. So if any factorisation exists at all, the formula $\bar\varphi(gN) = \varphi(g)$ is the only candidate. We take this formula as the definition and check that it produces a function — that is, that the value assigned to $gN$ does not depend on which representative $g$ we picked out of the coset. Suppose $gN = g'N$, i.e. $g$ and $g'$ are two representatives of the same coset. The standard characterisation of equal left cosets says $gN = g'N$ if and only if $g^{-1}g' \in N$. So $g^{-1}g' \in N$, and by the hypothesis $N \subset \ker \varphi$, we get $g^{-1}g' \in \ker\varphi$, that is $\varphi(g^{-1}g') = e_H$. Applying that $\varphi$ is a homomorphism: \begin{align*} e_H = \varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g'), \end{align*} which rearranges to $\varphi(g) = \varphi(g')$. This is exactly where the hypothesis $N \subset \ker\varphi$ does its work. Without it, two different representatives of the same coset could land at different points of $H$, and the rule would not define a function. The condition $N \subset \ker\varphi$ is precisely the obstruction to representative-dependence. [/guided] [/step] [step:Verify $\bar\varphi$ is a group homomorphism] Let $aN, bN \in G/N$. By the [Quotient Group](/theorems/790), the operation on $G/N$ is $(aN)(bN) = (ab)N$. Then \begin{align*} \bar\varphi\bigl((aN)(bN)\bigr) &= \bar\varphi\bigl((ab)N\bigr) = \varphi(ab) = \varphi(a)\varphi(b) = \bar\varphi(aN)\,\bar\varphi(bN), \end{align*} where the third equality uses that $\varphi$ is a homomorphism. Hence $\bar\varphi$ is a group homomorphism. [/step] [step:Verify the factorisation identity $\bar\varphi \circ \pi = \varphi$] For any $g \in G$, \begin{align*} (\bar\varphi \circ \pi)(g) &= \bar\varphi(\pi(g)) = \bar\varphi(gN) = \varphi(g), \end{align*} using the definition of $\pi$ and the definition of $\bar\varphi$. Hence $\bar\varphi \circ \pi = \varphi$. [/step] [step:Show that any factorising homomorphism must equal $\bar\varphi$] Suppose $\psi: G/N \to H$ is any group homomorphism with $\psi \circ \pi = \varphi$. For every coset $gN \in G/N$, \begin{align*} \psi(gN) &= \psi(\pi(g)) = (\psi \circ \pi)(g) = \varphi(g) = \bar\varphi(gN). \end{align*} Since $\psi$ and $\bar\varphi$ agree on every element of $G/N$, $\psi = \bar\varphi$. This proves uniqueness. [/step] [step:Characterise injectivity of $\bar\varphi$ via $\ker\varphi = N$] A group homomorphism is injective if and only if its kernel is the [trivial subgroup](/page/Subgroup); for $\bar\varphi$ this means $\ker\bar\varphi = \{N\}$, where $N$ is the identity coset of $G/N$. We compute the kernel: \begin{align*} \ker\bar\varphi &= \{gN \in G/N : \bar\varphi(gN) = e_H\} = \{gN : \varphi(g) = e_H\} = \{gN : g \in \ker\varphi\}. \end{align*} Equivalently, $\ker\bar\varphi = \pi(\ker\varphi)$. Suppose first that $\ker\varphi = N$. Then $\ker\bar\varphi = \{gN : g \in N\} = \{N\}$ (since every $g \in N$ satisfies $gN = N$, by Step "Identify the kernel as $N$" of [Properties of the Canonical Projection](/theorems/2700)). So $\bar\varphi$ is injective. Conversely, suppose $\bar\varphi$ is injective, so $\ker\bar\varphi = \{N\}$. The inclusion $N \subset \ker\varphi$ holds by hypothesis. For the reverse inclusion, take $g \in \ker\varphi$; then $\bar\varphi(gN) = \varphi(g) = e_H$, so $gN \in \ker\bar\varphi = \{N\}$, giving $gN = N$ and hence $g \in N$. Therefore $\ker\varphi \subseteq N$, and combined with the hypothesis we obtain $\ker\varphi = N$. [/step] [step:Characterise surjectivity of $\bar\varphi$ via surjectivity of $\varphi$] We compute the image: \begin{align*} \operatorname{im}\bar\varphi &= \{\bar\varphi(gN) : g \in G\} = \{\varphi(g) : g \in G\} = \operatorname{im}\varphi, \end{align*} using the definition of $\bar\varphi$ and the fact that every coset has the form $gN$ for some $g \in G$. Hence $\bar\varphi$ is surjective (i.e. $\operatorname{im}\bar\varphi = H$) if and only if $\operatorname{im}\varphi = H$, which is exactly the surjectivity of $\varphi$. This completes the proof of all four assertions: existence and uniqueness of $\bar\varphi$, the explicit formula $\bar\varphi(gN) = \varphi(g)$, the injectivity criterion $\ker\varphi = N$, and the surjectivity criterion. [/step]