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This question is related to this one.

Consider a function $g:\mathbb{R}\rightarrow \mathbb{R}$ and its left-inverse $f$ $$\forall x \in \mathbb{R} : f(g(x))=x $$ Then $$g(f(x))=x $$ has a set of solutions (usually a set with the same cardinality as $\mathbb{R}$. For example, this answer constructs $f$ and $g$ such that the solution set is an interval, and if $f$ is a true inverse of $g$ then the solution set is all of $\mathbb{R}$). But how few solutions can it have? Is there some $f,g$ for which it has countably many solutions? Finitely many? One? None?

Wouter
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Whenever $x$ lies in the image of $g$, it must satisfy $g(f(x)) = x$, since if $x=g(y)$, then

$$g(f(x)) = g(f(g(y)) = g(y) = x.$$

However if $g$ has a left inverse $f$ it must be injective:

$$g(x) = g(y) \implies f(g(x)) = f(g(y)) \implies x=y.$$

Therefore $g$ is a bijection from $\mathbb{R}$ to its image $g(\mathbb{R})$, and so the solution set $S=\{x\in\mathbb{R} : g(f(x)) = x\}$, which satisfies $g(\mathbb{R})\subseteq S\subseteq\mathbb{R}$, has the same cardinality as $\mathbb{R}$.

Peter Huxford
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