Do you agree that my proof of the following problem is correct? Due perhaps due to my limited knowledge, my perplexity arises from the fact that when $S$ is injective, it has almost the same kernel as $Q$.
Problem. Let $T:V\to W$ be a linear map, $M$ a linear subspace of $V$ s.t. $M\subset \ker T$, and $Q:V\to V/M$ the quotient map. Prove:
- There exists a unique linear map $S:V/M\to W$ s.t. $T=SQ$.
- $S$ is injective if and only if $M=\ker T$.
It's understood that $V$ and $W$ are built upon the same scalar field.
Proof of 1. I'm pretty sure it's fine.
Proof of 2.
$\Rightarrow )$ Already $M\subset \ker T$. $S$ injective means that $\ker S=\{u\in V/M:Su=\theta \}$ equals $\{M\}$, the set containing the zero vector of $V/M$. Let $x\in \ker T$, then $Tx=\theta$. Let $Qx=x+M=u$, then $Su=S(x+M)=Tx=\theta$, thus $x+M\in \ker S=\{M\}$, so $x+M=M$, hence $x\in M$.
$\Leftarrow )$ Let $M=\ker T$ and consider $u\in \ker S$. We show that $u\in \{M\}$. Now $u=x+M$ so $Su=Tx=\theta$, thus $x\in M$, therefore $u=x+M=M$, so $u\in \{M\}$. Notice that already $\{M\}\subset \ker S$ since $\ker S$ is a linear subspace of $V/M$.
Your proof is fine. There's actually a slightly more general result you can use to derive $\implies$ in $(2)$. Let $U,V,W$ be vector spaces and let $T: U \to V$ & $S: V \to W$ be linear maps. If $S$ is injective, then: $$\ker(S \circ T) = \ker(T)$$
The proof of this is quite easy (ask if you want me to write it down explicitly). In your case, observe that when $S$ is injective, then:
$$\ker(SQ) = \ker(Q) = \ker(T) = M$$
So, this isn't something that's just a very specific thing. It does arise from more general considerations and a more general statement about linear maps.