Prove that if $S: U\rightarrow V$ and $T: V\rightarrow W$ are isomorphisms, then $TS$ is also an isomorphism?

Question

I've already figured out that TS must be injective, but I'm struggling to show how TS is surjective. I have a sense of why this is true but I can't quite put it into words. Could someone help me out with this?

Also what is the inverse of TS?

Answer 1

Suppose $U$ and $V$ are groups and $S:U\to V$ and $T:V\to W$ are isomorphisms. If $u,u'\in U$ with $ST(u) = ST(u')$, then $Tu = Tu'$ since $S$ is injective, and hence $u = u'$ since $T$ is injective. If $w\in W$, then because $T$ is surjective, there exists $v\in V$ with $Tv=w$, and similarly there exists $u\in U$ with $Su=v$ since $S$ is surjective, and therefore $TS(u)=w$.

If $u,u'\in U$ then $S(Tuu')=S(Tu)(Tu')$ because $T$ is a homomorphism, and similarly $S(Tu)(Tu') = (ST(u))(S(Tu'))$ because $S$ is a homomorphism. It follows that $ST$ is an isomorphism.

Answer 2

The easiest and the fastest way to prove that $T\circ S$ is an isomorphism is by constructing the inverse. Put

$$F:=S^{-1}\circ T^{-1}$$

It is extremely easy to check (by composing $F\circ T\circ S$ and $T\circ S\circ F$) that $F=(T\circ S)^{-1}$.