\b[False.] A <left inverse> would force $\alpha$ to be <injective>: if $\alpha u=0$, then $u=\beta\alpha u=0$. Take the <surjective linear map> $\alpha:\mathbb R^2\to\mathbb R$ given by $\alpha(x,y)=x$. Since $(0,1)$ belongs to its <kernel>, every composite $\beta\alpha$ annihilates $(0,1)$ and cannot be the <identity map> on $\mathbb R^2$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 9F / i / Solution

{title2=$g\circ f=\operatorname{id}_X$}

A left inverse of a <function> $f:X\to Y$ is a function $g:Y\to X$ satisfying $g\circ f=\operatorname{id}_X$. It forces $f$ to be <injective>, since equal images can be mapped back to equal original elements.

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