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

Question

So if $S$ and $T$ are isomorphisms, then there exists $T^{-1}$ and $S^{-1}$, and then there exists $T^{-1}: W\rightarrow V$ and $S^{-1}: V\rightarrow U$, and so there exists $S^{-1} T^{-1}= (TS)^{-1}$. Could someone fix my proof?

Answer 1

I think is better to work with the definition of an isomorphism being one-to-one and onto.

Let's prove that $TS$ is one-to-one.

Let $u_{1}$,$u_{2}\in U$. Suppose that

$$ TS(u_{1}) = TS(u_{2})\\ T(S(u_{1}) = T(S(u_{2}) $$ Bu by definition T is one-to-one (isomorphism). Then: $$ S(u_{1}) = S(u_{2}) $$ But S is also an isomorphism $\implies$ $S$ is one-to-one, thus: $$ u_{1} = u_{2} $$ We can finally conclude that $TS$ is one-to-one.

Try using the same thought to prove that $TS$ is onto. Also, linearity of $TS$ is not difficult.