Let $\psi:G \rightarrow H$ hom, $K = \ker(\psi)$, and $\varphi$ be canonical projection. Show that $\psi(x)=\psi(y) \iff \varphi(x)=\varphi(y)$.

Question

Question :
Let $\psi : G \rightarrow H$ be a homomorphism, and set $K = \operatorname{ker}(\psi)$. Let $\varphi : G \rightarrow G/K$ be the canonical homomorphism defined by $\varphi(x) = xK$.

Show that $\psi(x) = \psi(y)$ $\iff$ $\varphi(x) = \varphi(y)$ (Hint: The First Isomorphism Theorem should be helpful here.)

I am not sure how to proceed here since $\psi$ is not defined. they just stated the domain and the codomain, but not the actual function like they did with $\varphi$. Thus after starting with $\psi(x) = \psi(y)$, I don't know what to do since $\psi$ isn't defined anywhere.

Answer 1

The reverse direction is fairly straightforward. Notice if $\varphi$ is the canonical projection map defined by $g \mapsto gK$ then $\varphi(x) = \varphi(y)$ means that $xK =yK$ or that $xy^{-1} \in K = \ker(\psi)$. If $xy^{-1}$ lives in the kernel of $\psi$, then you know $\psi(xy^{-1}) = 1$ so that by group homomorphism properties $$ 1= \psi(xy^{-1}) = \psi(x)\psi(y^{-1})= \psi(x)\psi(y)^{-1} \implies \psi(x) = \psi(y). $$

For the other direction, the first isomorphism theorem tells you that $G/K \cong \psi(G)$ so there is a natural isomorphism, say $\rho:G/K \rightarrow \psi(G)$, between the quotient by the kernel and the image of your homomorphism. Moreover, it tells you the group homomorphism $\psi$ factors via $$ \psi(g) = (\rho\circ \varphi)(g). $$ Notice that $\psi(x),\psi(y)$ are certainly elements in the image of $\psi$ so that $\psi(x) = (\rho \circ \varphi)(x)$ and $\psi(y) = (\rho \circ \varphi(y))$. Therefore $\psi(x) = \rho(xK)$ and $\psi(y) = \rho(yK)$. If these are assumed to be equal then $$ \rho(xK) = \rho(yK), $$ but since $\rho$ is an isomorphism of groups it possesses an inverse so we can conclude $xK =yK$ which implies that $xy^{-1} \in K$ i.e $\varphi(xy^{-1}) = 1_{G/K}$ so $\varphi(x) = \varphi(y)$.

Edit: Not to belabour the point, but I just read your comment to Burde, your map $f(xK) = \psi(x)$ is exactly the map I've denoted $\rho$ which is guaranteed to exist by the first isomorphism theorem. For reference

Theorem: Let $\varphi:G \rightarrow H$ be a homomorphism, $N \unlhd G$ and $\pi:G \rightarrow G/N$ be the natural projection map. Then there is a well-defined homomorphism $\psi:G/N \rightarrow \varphi(G)$ such that $\varphi = \psi \circ \pi$ if and only if $N \subseteq \ker(\varphi)$. If this is the case $\psi$ is unique and given by $\psi(gN)=\varphi(g)$.

With reference to the theorem, when you quotient by the kernel; i.e $N = \ker(\varphi)$, the map $\psi$ is an isomorphism showing $G/N \cong \varphi(G)$.