I am unable to prove injectivity for the first isomorphic theorem. A website shows the proof as follows: Map: $ϕ(xN) = ψ(x)$ ϕ is injective:
$ϕ(xN)=ϕ(yN)\iff ψ(x) = ψ(y)\iff eH = ψ(x)^{-1} ψ(y) = ψ(x^{-1}) ψ(y) = ψ(x^{-1}y)\iff x^{-1}y\in N \iff xN = yN$
I don't understand how $x^{−1}y ∈ N ⇐⇒ xN = yN$ imply that $\phi$ is injective?
Also, is there an alternative way to show that $\phi$ is injective by concluding that x=y?
Note: N here denotes the kernel of $ψ$
On
It seems your trouble is in understanding that if $N$ is a normal subgroup, then $$xN=yN \iff x^{-1}y \in N.$$
Let $xN=yN.$ Then $N=x^{-1}yN.$ Since, $N$ is a subgroup, therefore it contains the identity, say $e.$ Thus, $(x^{-1}y)e \in x^{-1}yN=N.$ So, $x^{-1}y \in N.$
Conversely, let $x^{-1}y \in N.$ Again, since $N$ is a subgroup therefore, $x^{-1}yN \subseteq N.$ Thus, $yN\subseteq xN.$ Since, two cosets are either disjoint or identical, therefore $xN=yN.$
You have $$ϕ(xN) = ϕ(yN)\Longrightarrow xN = yN.$$ So $\phi$ is injective.