Proof verification: Linear maps

Question

Question: Let $S:U \longrightarrow V$ and $T: V \longrightarrow W$ be linear maps. Prove that $\operatorname{Range}(TS) \subseteq \operatorname{Range}(T)$.

Proof: Let $u \in U$, $v \in V$, $w \in W$ be some vectors. Suppose that $$w = T(S(u)) = T(v).$$ This means that $w \in \operatorname{Range}(TS)$, since $w$ is in the image of the composition $TS$. But then $w$ must be in the image of $T$, so $w \in \operatorname{Range}(T)$. Since $w \in \operatorname{Range}(TS)$ and also $w \in \operatorname{Range}(T)$, we have that $\operatorname{Range}(TS) \subseteq \operatorname{Range}(T)$.

(I am unsure about this proof since I didn't rely on the fact that $S$ and $T$ are linear maps and I don't know if my choice of vectors applies generally.)

Answer 1

Yes, it has nothing to do with linearity. This holds for any such mappings.

Start of the proof is not correct.

To show $\operatorname{Range}(TS)\subseteq\operatorname{Range}(T)$, start with letting $w\in W$ such that $w\in \operatorname{Range}(TS)$. Then $w=TS(u)$ for some $u\in U$. Then $w=T(S(u))\implies w\in\operatorname{Range}(T)$. Thus it is proved.

Answer 2

I'd adjust the last sentence to read "when $w\in\operatorname{Range}TS$, then also $w\in\operatorname{Range} T$. Also, $v,w$ are not arbitrary. They depend on your choice of $u$. The point is that if $w$ is the image of an arbitrary vector $u$ under $TS$, then it's also the image of the (not arbitrary) vector $S(u)$ under $T$. And you're right, the claim is true for arbitrary maps on arbitrary sets. Them being linear maps isn't necessary at all.

Answer 3

Whenever you prove this type of inclusion then enumerate an element from 'left' and show that this is also in 'right'. That will work for arbitrarily elements.

Suppose $w\in \operatorname{Range}(TS)$. Then $TS(u)=w$ for some $u$ in $U$. $TS(u)=T(S(u))=T(v)=w$ for some $v\in V$.i.e, $w\in \operatorname{Range}(T)$. It follows that $\operatorname{Range}(TS)\subseteq \operatorname{Range}(T)$.