Let $T: V \rightarrow W$ and $S: W \rightarrow Z$ be linear maps. If $S\circ T$ is an isomorphism, are $S$ and $T$ isomorphisms? 
Let $V$, $W$, and $Z$ be finite-dimensional vector spaces.  Let $T: V \rightarrow W$ and $S: W \rightarrow Z$ be linear mpas. If $S\circ T$ is an isomorphism, then $T$ and $S$ are isomorphisms.

I am told that this is in general not true, but I am not sure where to start to prove untrue. Can anyone give references to explore on this?
 A: One counterexample I can give is 
$T:\begin{cases}R \rightarrow R^2 \\ x \rightarrow (x,0)\end{cases}$
$U:\begin{cases} R^2 \rightarrow R \\ (a,b) \rightarrow a \end{cases}$
UT is an isomorphism while T, S are not.
A: Let $V$, $W$, and $Z$ be vector spaces (not assumed to be finite dim) over some skew field $D$ (such as $D=\Bbb{F}_{p^k}$, $D=\Bbb Q$, $D=\Bbb R$, $D=\Bbb C$, or $D=\Bbb H$ the quaternions).  Let $T:V\to W$ and $S:W\to Z$ be linear maps.  
(a) Then, $S\circ T:V\to Z$ is an isomorphism of vector spaces if and only if all conditions below are met:


*

*$T$ is injective,

*$S$ is surjective, and

*$W=\operatorname{im}T\oplus \ker S$.


(b) The three conditions above can be fulfilled if and only if $$\dim V=\dim Z\le \dim W.$$
(c) Let $S\circ T:V\to Z$ be an isomorphism.  The following are equivalent:


*

*$T$ is an isomorphism,

*$T$ is surjective,

*$S$ is an isomorphism, and

*$S$ is injective.



Proof of (a)
$(\Rightarrow)$  Since $S\circ T$ is an isomorphism, it is both injective and surjective.  Therefore, $$\ker T\subseteq \ker S\circ T=0$$
making $T$ injective, and $$\operatorname{im}S\supseteq \operatorname{im}S\circ T=Z$$ making $S$ surjective.  If $w\in \operatorname{im}T\cap\ker S$, then $Sw=0$ and there exists $v\in V$ s.t. $Tv=w$.  Thus, $(S\circ T)v=S(Tv)=Sw=0$.  As $S\circ T$ is an isomorphism, $v=0$.  Therefore, $w=0$ so $$\operatorname{im}T\cap\ker S=0.$$  We define a a map $P:W\to W$ via
$$P=T\circ (S\circ T)^{-1}\circ S.$$
We clearly have $\operatorname{im}P\in \operatorname{im} T$.  Furthermore,
\begin{align}P^2 &=\big(T\circ(S\circ T)^{-1}\circ S\big)\circ \big(T\circ(S\circ T)^{-1}\circ S\big)\\&=T\circ \big((S\circ T)^{-1}\circ (S\circ T)\big)\circ (S\circ T)^{-1}\circ S\\&=T\circ (S\circ T)^{-1}\circ S=P,\end{align}
so $P$ is a projection.  If $\text{id}_X:X\to X$ is the identity map of a vector space $X$, then
\begin{align}S\circ(\text{id}_W-P)&=S-S\circ P= S-S\circ\big(T\circ (S\circ T)^{-1}\circ S\big)\\&=S-\big((S\circ T)\circ (S\circ T)^{-1}\big)\circ S =S-S=0.\end{align}
Therefore, $\operatorname{im}(\text{id}_W-P)\subseteq \ker S$.  This means
$$w=Pw+(\text{id}_W-P)w \in \operatorname{im}T+\ker S$$
for all $w\in W$.  Thus, $W= \operatorname{im} T+\ker S$.  Since $\operatorname{im} T\cap \ker S=0$, $W=\operatorname{im}T\oplus \ker S$, as claimed.
$(\Leftarrow)$ Let $\pi:W\to \operatorname{im} T$ be the projection given by the decomposition $W=\operatorname{im}T\oplus\ker S$.  Define a map $\sigma:Z\to \operatorname{im} T$ as follows: for $z\in Z$, by surjectivitiy of $S$ there exists $w \in W$ such that $Sw=z$, and then we set $$\sigma z=\pi w.$$  We claim that this map is well defined.  If $w,w'\in W$ are s.t. $Sw=z$ and $Sw'=z$, then $w-w'\in\ker S$; therefore $$\pi w-\pi w'=\pi(w-w')=0,$$ making $\pi w=\pi w'$.  Now because $T$ is injective, $\pi\circ T:V\to \operatorname{im} T$ is an isomorphism.  Let $\tau:\operatorname{im} T\to V$ be the inverse of $\pi\circ T$.  We consider $\rho:Z\to V$ to be the composition $\tau\circ \sigma$.  Observe that $$\sigma \circ S=\pi$$
and $$T\circ\tau =\iota,$$
where $\iota:\operatorname{im}T\to W$ is the inclusion map.  Thus
$$(S\circ T)\circ \rho=S\circ (T\circ \tau)\circ \sigma=S\circ \iota\circ\sigma=\operatorname{id}_Z$$
and
$$\rho\circ(S\circ T)=\tau\circ(\sigma\circ S)\circ T=\tau\circ \pi\circ T=\operatorname{id}_V.$$
Hence $\rho:Z\to V$ is the inverse of $S\circ T:V\to Z$.

Proof of (b)
$(\Rightarrow)$  If the three conditions are fulfilled, then there exist such maps $S$ and $T$ with $S\circ T$ being an isomorphism from $V$ to $Z$.  Thus, $V\cong Z$, making $\dim V=\dim Z$.  But since $W=\operatorname{im} T\oplus \ker S$, $$\dim W=\dim\operatorname{im} T+\dim\ker S.$$  As $T$ is injective, $V\cong \operatorname{im}T$, so $\dim\operatorname{im} T=\dim V$.  This shows that
$$\dim W=\dim V+\dim\ker S\ge \dim V.$$
$(\Leftarrow)$  Suppose that we have $\dim V=\dim Z\le \dim W$.  The claim is trivial if $\dim V=0$.  We now assume that $\dim V>0$.  Let $\mathcal{V}$, $\mathcal{W}$, and $\mathcal{Z}$ be bases of $V$, $W$, and $Z$, respectively. Since $\dim V\le \dim W$, there exists an injection $f:\mathcal{V}\to\mathcal{W}$.  Since $\dim V=\dim Z$, there exists a bijection $h:\mathcal{V}\to\mathcal{Z}$.  Since $\dim Z=\dim V>0$, $\mathcal{Z}\ne\emptyset$, so it contains an element $z_0$.  We define a surjection $g:\mathcal{W}\to\mathcal{Z}$ as follows:
$$g(w)=\left\{\begin{array}{ll}
h(v)&\text{if}\ w=f(v)\ \text{ for some }v\in\mathcal{V},\\
z_0&\text{otherwise}. 
\end{array}
\right.$$
Let $T:V\to W$ be the linear map that linearly extends $f$, and $S: W\to Z$ be the linear map that linearly extends $g$.  Then, $T$ and $S$ satisfy the three conditions.

Proof of (c)
If $T$ is an isomorphism, then it clearly is surjective.  If $T$ is surjective, then from $W=\operatorname{im}T\oplus\ker S$, we must have $\ker S=0$, so $S$ is injective.  If $S$ is injective, then as $S$ is surjective, $S$ is an isomorphism.  If $S$ is an isomorphism, then since $S\circ T$ is an isomorphism, $T=S^{-1}\circ(S\circ T)$ is an isomorphism.
Postscript:  Any example of the pair $S:V\to W$ and $T:W\to Z$ s.t. $S\circ T:V\to Z$ is an isomorphism but $S$ and $T$ are not isomorphisms can be constructed as follows.  Let $V$ and $Z$ be two isomorphic vector spaces.  Take $R:V\to Z$ to be an isomorphism.  Pick an arbitrary vector space $W$ with an injective map $T:V\to W$ which is not an isomorphism.  (If $V$ is finite dim, then this can be done by choosing any vector space $W$ with a larger dimension.)  Then $W=\operatorname{im} T\oplus K$ for some non-zero subspace $K$ of $W$.   (Alternatively, let $V'$ be another vector space isomorphic to $V$ with an isomorphism $T':V\to V'$, and $K$ a non-zero vector space.  Set $W=V'\oplus K$, $T:V\to W$ by extending $T'$ trivially.)  Let $\varpi:W\to\operatorname{im} T$ be the projection induced by the decomposition $W=\operatorname{im} T\oplus K$. Define $S:W\to Z$ by sending $w\mapsto Rv$ if $v\in V$ is the unique element s.t. $Tv=\varpi w$.
