Exercise I.B.3

Question

Let $V,W$ be vector spaces, and let $\phi:V	o W$, $\varphi:W	o V$ be a linear transformation such that $\varphi\circ \phi: V	o V$ is an isomorphism. Show that $\phi$ is injective and $\varphi$ is surjective.

Toğrul Topal · Accepted Answer

We argue by contradiction.
\begin{itemize}
\item Suppose that $\phi$ is not injective, i.e., there exists a $w \in W$ for which there are at least two distinct $v,v' \in V$ with $\phi(v) = \phi(v') = w$. Let $z := \varphi(w)$. But then we have both $v,v' \in (\varphi \circ \phi)^{-1}(\{z\}) = \phi^{-1}\varphi^{-1}(\{z\})$ - a contradiction to the bijectivity of the composition.
\item Suppose that $\varphi$ is not surjective. Then there exists $y\in V$ such that there is not $w\in W$ with $\varphi(w) =y$. But then we cannot take the inverse of $\varphi \circ \phi$ under $y$ - a contradiction.
\end{itemize}

Exercise I.B.3

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Exercise I.B.3

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