In my textbook the following is proposed as a theorem (with some explanation but without proof), but it seems wrong to me. Maybe it's an easily-fixable typo; I don't know.
I will quote the (so-called) theorem and then point out its error. You may want to inform me if I'm wrong or mistaken (I'm prone to make mistakes all the time!), and tell me if you were the editor, what would be the best thing to change about this theorem to correct it?
Let $V$ and $W$ be vector spaces and $T: V→W$ a linear map. Let $U$ be a subspace of $V$. Define $T':V/U→T(V)$ by $T'[a]=Ta$, where $[a]:= \left\{a+u :u\in U\right\}$. Then $T'$ is a well-defined isomorphism. (i.e. a bijective linear map, a one-to-one linear correspondence)
Why it's wrong:
First we have to check if $T'$ is a well-defined linear map, that is if $[a]=[b]$ (with $b$ being another representative for the set $[a]$) then $T'[a]=T'[b]$, which by definition would lead to $Ta=Tb$. But this is not generally true unless $a-b \in ker T$, in other words, unless $U = kerT$.
But this is not a good correction, because it reduces this theorem to a prevoius example: just a repetition.
We need, that $a-b\in U \Rightarrow Ta = Tb$. If that is the case, then if $[a]=[b]$ we have $a - b\in U$, so $Ta = Tb$, hence $T'[a] = Ta = Tb = T'[b]$.
The property $a-b\in U \Rightarrow Ta = Tb$ is equivalent to $U\subseteq \ker T$:
Assume $a-b\in U \Rightarrow Ta = Tb$ and let $u\in U$. Then $u-0\in U$, so $Tu = T0 = 0$, so $u\in \ker T$. Hence $U\subseteq \ker T$.
On the other hand assume $U\subseteq \ker T$ and let $a-b\in U \subseteq \ker T$. Then $Ta-Tb = T(a-b) = 0$, so $Ta = Tb$. Hence the property: $a-b\in U \Rightarrow Ta = Tb$.
To fix the theorem, we indeed need $U\subseteq \ker T$ (but not necessarily $U\supseteq \ker T$).