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Need some hints to solve Ex6a from V. Zorich course of Analysis vol.1 chap.1 §3.
If mappings $f:X\to Y$ and $g:Y\to X$ are such that $g \circ f=id_X$ where $id_X$ - identity map X, then $g$ is called left inverse for $f$ and $f$ is called right inverse for g. Show that unlike the unique inverse map, there can exist many one-sided inverse maps.

Author's hint to regard set $X$, for example consisting of one element, set $Y$ of two elements.

My thoughts: using the fact that $g(f(x))=x$ and $f(g(y))=y$, then (may be) to use
lemma: $g \circ f=id_X => (g \quad surjective)\wedge(f\quad injective)$

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1 Answer 1

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Your lemma says it all. For example, take $g(x,y,z)=x$ (from $\mathbb{R}^3$ to $\mathbb{R}$). We see that it is surjective. Now any map $f:\mathbb{R}\to\mathbb{R}^3$ that sends $x$ to something with first coordinate $x$ works. For example, $f(x)=(x,h(x),k(x))$ works for any functions $h$ and $k$.

Note: Here I randomly chose $\mathbb{R}$ and $\mathbb{R}^3$; try to generalize this!

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  • $\begingroup$ Is it enough to state: 1) $im(g)=X$ - surjectivity of g and, therefore many then 2) define (then proof?) $id_y=f \circ g\rightarrow(f \quad surjective)\wedge(g \quad injective)$ ? $\endgroup$
    – Arteom.k
    Sep 28, 2014 at 16:03
  • $\begingroup$ Dear @Arteom.k, I'm not sure I understand what you're asking. $\endgroup$
    – rfauffar
    Sep 28, 2014 at 19:23

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