$f:R\to \text{Im}\:f,\:\pi : R\to R/\ker f,\;f=g\circ \pi$. Check if $g:\:R/\ker f\to \text{Im}\:f$ is an isomorphism

Question

$\DeclareMathOperator{\Im}{Im}$ I can't really think of a way to prove it, I know that we have to prove that

$$g\bigl([x_1]\bigr)=g\bigl([x_2]\bigr)\:\Rightarrow [x_1]=[x_2]$$

And

$$\forall y\in \Im f\enspace\exists x \in R/\ker f,\;y=g(xt)$$

But I don't really know how to start.

Answer 1

definition of $g$

we define $g$ as \[ g \colon R / \ker f \ni [x_1] \mapsto f(x_1) \in \operatorname{Im}f . \] We have to show that this definition doesn't depend on representation, i.e. if $[x_1] = [x_2]$, then $g([x_1]) = g([x_2])$. In other words, if $x_1 - x_2 \in \ker f$, show that $f(x_1) = f(x_2)$, which follows from definition of a kernel. This is what is called well-definedness, and often matters when there is quotient.

surjectivity

Let us show surjectivity:

for all $y \in \operatorname{Im}f$, is there $[x] \in R / \ker f$ such that $g([x]) = y$ ?

Since $y \in \ker f$, there is $x' \in R$ such that $f(x') = y$. Does $\pi(x') = [x']$ suit for $[x]$ mentioned above ?

injectivity

Now, injectivity:

Let $[x_1], [x_2] \in R / \ker f$ and $g([x_1]) = g([x_2])$. To show that $g$ is injective, it is sufficient to show that $[x_1] = [x_2]$.

Since $g([x_1]) = g([x_2]) = f(x_1) = f(x_2)$, $f(x_1) - f(x_2) = f(x_1 - x_2) = 0$, which means $x_1 - x_2 \in \ker f$. Hence $[x_1] = [x_2] \in R/ \ker f$.

Of course, it is ok to show that $\ker g = \{0\}$. This is a good exercise.