Proving that $\operatorname{Im} f \cong V/\ker f$ for a linear map f.

Question

In this proof, how do you show injectivity? In my notes, it says:

For $f\colon V \to W$ linear operator and $V,W$ vector spaces let $\bar{f}\colon V/\ker f \to \operatorname{Im}f$ be defined by $\bar{f}(v+\ker f) = f(v)$. If $\bar{f}(v+\ker f)=0 $, we have that $ v \in \ker f $ so $v+\ker f = 0 + \ker f \implies \ker \bar{f} = \{ 0 + \ker f \} $ and so $\bar{f}$ is injective.

How is the step $ v \in \ker f \implies v+\ker f = 0 + \ker f $ obtained? Am I missing something obvious?

Answer 1

Both $v+\ker f$ and $0+\ker f$ are cosets. In fact, they are the same coset. The quick way to check this is to use the theorem that if $W$ is a subspace, then $a+W = b+W$ if and only if $a-b \in W$. In this case, $v-0 \in \ker f$ so $v+\ker f= 0 + \ker f$.

Answer 2

$\ker f$ is a vector space. Therefore, for all $x,y\in \ker f$, $$x+y\in \ker f.$$

In particular $v\in \ker f$, therefore $v+\ker f\underset{(*)}{=}\ker f=0+\ker f$

$(*)$ is abusive notation, but it's to give you the idea.