Homomorphism is Injective if and only if the kernel is trivial

Short, sweet, to the point.

Injective definition:

Injective means each output has a distinct input. Mathematically we can check for this by saying $f(a)=f(b)⟺a=b$ for some map f

Proof:

Let there be some homomorphism from $\varphi :G\to K$. Then $\varphi$ is injective $⟺$ $Ker(\varphi )=\{e\}$.

1. $\Rightarrow$
   1. Assume $\varphi$ is injective.
      1. Then, if there is some $x\in Ker(\varphi )$ we have it so $\varphi (x)=e$.
      2. However, we also know $\varphi (e)=e$. As this is injective, we know $x=e$ and the kernel is trivial.
2. $\Leftarrow$
   1. Assume $Ker(\varphi )=\{e\}$.
      1. Then, to show injective assume $\varphi (x)=\varphi (y)$: $\varphi (x)=\varphi (y)⟹\varphi (x)\varphi (y{)}^{-1}=e⟹\varphi (x{y}^{-1})=e$
      2. However, we know that the only thing in the kernel is e, so we know only $\varphi (e)=e$. So therefore: $x{y}^{-1}=e⟹x=y$
      3. We have proved injective as $\varphi (x)=\varphi (y)⟹x=y$