How do I show that g ◦ f = id, when I only have the information that f is injective?

Asked Dec 05 '23 at 14:28

Active Dec 05 '23 at 14:28

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Let f ∈ Hom(R3,R7) be an injective map, then there exists a g ∈ Hom(im(f), R3 ) such that g ◦ f = id.

We have that invertibility is equivalent to injectivity and surjectivity. But how do I show that g ◦ f = id, when I only have the information that f is injective?

asked Dec 05 '23 at 14:28

Marie

A map of sets $f:X\to Y$ is injective if and only if there is some $g: f(X)\to X$ with $g \circ f = \text{id}_X$. – psl2Z Dec 05 '23 at 14:33
What do you mean by R3 and R7? – ShyamalSayak Dec 05 '23 at 14:41
I meant ℝ^3 and ℝ^7 – Marie Dec 05 '23 at 14:46
Define $g: \operatorname{Im}(f) \to \mathbb{R}^{3}$ as $g(y) = x$, where $f(x) = y$. Since $f$ is an injective function it follows that $g$ is well-defined. – ZAF Dec 05 '23 at 15:10

How do I show that g ◦ f = id, when I only have the information that f is injective?

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