Prove that $g:R/\ker(φ)\to \operatorname{Ιm}(\varphi)$ where $ g(\ker \varphi+r)=\varphi(r)$ is $1$-$1$ Given a ring homomorphism $\varphi:R\to S$, how to prove that:

$g:R/\ker(φ)\to \operatorname{Ιm}(\varphi)$ where $ g(\ker \varphi+r)=\varphi(r)$ is $1$-$1$

I know that function is an isomorphism but i can't prove the 1-1 property. I tried to prove that $\ker g$ contains only the $0$ element but i kinda stuck on the properties of the quotient ring maybe I'm not using them right.
 A: I think the key thing to remember here is that though $\ker \varphi$ is a subset of $R$, typically with more than one element, it is but a single element of $R/\ker \varphi$, that is, it is the zero element in the ring of cosets $\{r + \ker \varphi\}$, where $r \in R$.  Now if $g(r + \ker \varphi) = g(s + \ker \varphi)$, then we have $\varphi(r) = \varphi(s)$ since $g(r + \ker \varphi) = \varphi(r)$ for all $r \in R$.  $\varphi(r) = \varphi(s)$ implies $\varphi(r - s) = 0$, since $\varphi$ is a ring homomorphism.  Thus $r - s \in \ker \varphi$.  We now invoke the well-known property that, as cosets, $r + \ker \varphi = s + \ker \varphi$ if and only if $r - s \in \ker \varphi$.  Indeed, it is easy to see that this logical equivalence must be the case, for if $r + \ker \varphi = s + \ker \varphi$, we must have $r = r + 0 = s + k$ for some $k \in \ker \varphi$; then $r - s = k \in \ker \varphi$; going the other way is just as simple, for if $r - s \in \ker \varphi$ there is a $k \in \ker \varphi$ with $r - s = k$ or $r = s + k$, and if $t \in r + \ker \varphi$ the we have $t = r + l$ for some $l \in \ker \varphi$, whence $t = s + (k + l) \in s + \ker \varphi$ since both $k, l \in \ker \varphi$; this shows that $r + \ker \varphi \subseteq s + \ker \varphi$ and interchanging the roles of $r$ and $s$ yields $s + \ker \varphi \subseteq r + \ker \varphi$, whence these two cosets are equal:  $r + \ker \varphi = s + \ker \varphi$.  We thus see that $g: r + \ker \varphi \mapsto \varphi(r)$ maps precisely one coset $r + \ker \varphi$ to $\varphi(r)$, hence on the set of cosets $R/\ker \varphi$, $g$ is one-to-one.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
