How do I prove that the relation $x \sim y \Leftrightarrow f(x) = f(y)$ creates an injective function $X/\sim \to Y$? On page 127 of Analysis I by Amann and Escher, I am asked to prove that $E/\ker(T) \cong \text{im}(T)$ where $T \colon E \to F$ is a linear map between vector spaces over $K$. As far as I know this is a standard isomorphism theorem that is extremely similar to one of the group isomorphism theorems.
I am trying to understand what I perceive to be the fundamental reason those two isomorphism theorems (for vector spaces and groups) are true. I believe that fundamental reason is this excerpt from page 23 of my book:

.

Specifically, if you replace $Y$ with $\text{im}(f)$, then $\tilde{f}$ is a bijection. I think this is the essence of the two isomorphism theorems mentioned above, because formation of the quotient group $G/N$ is an equivalence relation on $G$ of the same sort, i.e. for $g, h \in G$, the equivalence relation $g \sim h \Leftrightarrow g \in h \odot N$ implies that $g \sim h \Leftrightarrow \varphi(g) = \varphi(h)$ where $\varphi \colon G \to G'$ is a homomorphism. Since a vector space is an abelian group with extra structure, the formation of a quotient vector space works in essentially the same way and gives rise to the same type of relations.
I think I understand most of what's going on, but I'm not 100% clear on why the function $\tilde f$ is injective. Intuitively this is obvious, because the projection map essentially gets rid of the "redundancy" of having multiple elements of the domain go to the same element in the codomain. Regarding the exercise requiring me to prove that $E/\ker(T) \cong \text{im}(T)$, I think that if I could just prove injectivity of $\tilde f$ in the diagram above, then injectivity of $\widetilde T \colon E/\ker(T) \to F$ would probably follow quickly.
As it is, if $\pi \colon E \to E/\ker{T}$ is my projection map, I end up arguing something like this, where $x, y \in E$:
\begin{align*}
\widetilde T\big(x + \ker(T)\big) &= \widetilde T\big(y + \ker(T)\big)\\
\Rightarrow T(x) &= T(y) \text{ (by definition of } \widetilde T)\\
\Rightarrow \widetilde T\big(\pi(x)\big) &= \widetilde T\big(\pi(y)\big)
\end{align*}
and then I'm stuck. I need to show that $x$ and $y$ are in the same coset. Alternatively I could show that if $x + \ker(T) \neq y + \ker(T)$, then $T\big(x + \ker(T)\big) \neq \widetilde T\big(y + \ker(T)\big)$. But I keep ending up thinking about $T(x) \neq T(y)$, which feels like begging the question.
Perhaps the proper way to show injectivity in the vector space case is to state that a linear map is a group homomorphism, so injectivity in the vector space case is taken care of by injectivity in the group case. But that still ultimately relies on injectivity in the excerpt from my text shown above, and I'm not quite sure how to prove that.
Maybe if I look at the excerpt above and argue as follows:
\begin{align*}
\tilde f \big([x]\big) &= \tilde f \big([y]\big)\\
\Rightarrow f(x) &= f(y)\\
\Rightarrow x &\sim y \text{ (by the equivalence relation)}\\
\Rightarrow [x] &= [y] \text{ (property of equivalence relations)}.
\end{align*}
Maybe that takes care of everything. I'm not 100% sure.
I appreciate any help.
 A: You seem to mix set theory with group (algebraic) theory. Let's straighten things up.
If $f:X\to Y$ is a function (on sets, no additional structure) then we have relationship $x\sim y$ iff $f(x)=f(y)$ on $X$. With this we define:
$$F:X/\sim\to Y$$
$$F([x])=f(x)$$

Lemma. $F$ is well defined and injective.

Proof. $[x]=[y]$ if and only if $x\sim y$ which is if and only if $f(x)=f(y)$. The "$\Rightarrow$" implication means $F$ is well defined. And "$\Leftarrow$" implication means it is injective. $\Box$
A simple corollary is that $F:X/\sim\to \text{im}(f)$ is a bijection.
Now if $G,H$ are groups and $f:G\to H$ is a group homomorphism then we have the kernel $\ker(f)=\{g\in G\ |\ f(g)=0\}$. We can now create the quotient group $G/\ker(f)$. In order to apply what we proved earlier all we need to know is that the quotient arises as $G/\sim$ from $x\sim y$ iff $f(x)=f(y)$. And that's straightforward:
$$x+\ker(f)=y+\ker(f)\text{ iff}$$
$$x-y+\ker(f)=\ker(f)\text{ iff}$$
$$x-y\in\ker(f)\text{ iff}$$
$$f(x-y)=0\text{ iff}$$
$$f(x)=f(y)$$
The last thing to prove is that the induced $F$ function is actually a homomorphism (which I leave as an exercise). With that we have the first isomorphism theorem.
