Show that the composition of isomorphisms is again an isomorphism

Let $U,V,W$ be vector spaces over $F$ and $S: U \to V$, and $T: V \to W$ linear maps.

(a) Show that if $S$ and $T$ are isomorphisms, then $T\circ S$ is an isomorphism, too.

(b) Show that if $U$ is isomorphic to $V$ and $V$ is isomorphic to $W$, then $U$ is isomorphic to $W$.

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In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Jun 8 '13 at 10:13

$a)$

$S\in GL(U,V), T\in GL(V,W)$

Since $GL(U,V) \subset L(U,V)$ and $GL(V,W) \subset L(V,W)$, $S\in L(U,V)$ and $t\in L(V,W)$ from which you can deduce $T\circ S \in L(U,W)$.

Then you must know that given two vector spaces $A$ and $B$, $\forall f \in L(A,B), [f\in GL(A,B) \Leftrightarrow f \text{ bijective}\Leftrightarrow f \text{ injective and surjective} \Leftrightarrow Ker(f)=\{0_A\} \text{ and } Im(f)=B]$

$\forall x \in U ,[(T\circ S)(x)=0_W \Leftrightarrow S(x)=T^{-1}(0_W) \Leftrightarrow S(x)=0_V \Leftrightarrow x=S^{-1}(0_V)\Leftrightarrow x = 0_U]$ which means that $\boxed{Ker(T\circ S) = \{0_U\}}$

Let $z\in W$

Since $T\in GL(V,W), \exists y \in V, T(y)=z$

Since $S\in GL(U,V), \exists x \in U, S(x)=y$

So $\forall z \in W, \exists x\in U, T(S(x))=z$ which means $W \subset Im(T\circ S)$

Since we already know that $Im(T\circ S)\subset W$,

$\boxed{Im(T\circ S)= W}$

So we get that $T\circ S\in GL(U,W)$

$b)$

Since $U$ and $V$ are isomorphic, $GL(U,V)\not = \emptyset$

Since $V$ and $W$ are isomorphic, $GL(V,W)\not = \emptyset$

So you can take $S\in GL(U,V)$ and $T\in GL(V,W)$

By $a)$, you get that $T\circ S \in GL(U,W)$

So $GL(U,W)\not = \emptyset$ so $U$ and $W$ are isomorphic.

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Hints:

• For (a): Check that the conditions on an isomorphism hold for $T\circ S$.

• For (b): Apply (a).

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