Let $V \xrightarrow{\phi} W \xrightarrow{\psi} V$. Show that $\phi$ is injective and $\psi$ is surjective.

Question

Let $V \xrightarrow{\phi} W \xrightarrow{\psi} V$ be linear maps such that $V\xrightarrow{\psi\phi}V$ is an isomorphism. Show that $\phi$ is injective and $\psi$ is surjective.

So, I know that an isomorphism is both injective and surjective. Surjective meaning that all of $W$ maps onto all of $V$ at least once and injective meaning that the nullspace of the map $V \rightarrow W$ will be only the zero vector.

But, how does one prove this without being given a function?

Answer 1

$\phi$ is injective Suppose $x,y\in V$ and $\phi(x) = \phi(y)$. Let $f = \psi\circ\phi$ be the isomorphism. Then $\psi(\phi(x)) = \psi(\phi(y)) = f(x)=f(y)$. As $f$ is invertible, $f^{-1}(f(x)) = f^{-1}(f(y)) = x = y$. A linear map is injective iff it's 1-1 iff its kernel is trivial.

$\psi$ is surjective If $\psi$ were not surjective, then $\psi\circ\phi$ would not be surjective — impossible, since that is a a bijection.

Answer 2

This results from a general fact about maps between sets:

Let $E\overset{f}{\longrightarrow} F\overset{g}{\longrightarrow}G$ be maps. Then

• if $g\circ f$ is injective, $f$ is injective.
• if $g\circ f$ is surjective, $g$ is surjective.

Indeed, if $f$ is not injective, $g\circ f$ can't be injective.

The second asserion results from $f(E)\subseteq F$, hence $g(f(E))\subseteq g(F)$, so if $g(f(E))=G$, then $g(F)=G$.