1
$\begingroup$

Let f : X → Y and g : Y → X be functions and assume $g ◦ f = I_X$. Prove of g is injective then $f ◦ g = I_Y$.

Approach if g is the left inverse of f then there exists $x\in X$ and $y \in Y$ such that $f(x)=y$ and $g(y)=x$.

I am confused about the left inverse of f. How is it possible that the left inverse of f is not its right inverse. If g is not one to one, this would imply that f is not a function or such g doesn't exist. I think that for $g ◦ f = I_X$ to be true g has to be injective.

In my approach, I think that a good start is to say that f(g(y))=y, but how can this be true? or how can this be false?

$\endgroup$
1
$\begingroup$

The key here is the fact that $g$ is injective (this is the crucial point, without that this will not go through).

Assume (i) $g\circ f = I_X$ and (ii) $g$ is injective. We want to establish $f\circ g = I_Y$, i.e. $f\circ g(y) = y$ for all $y\in Y$.

Take any $y\in Y$, and consider $z\stackrel{\rm def}{=} f\circ g(y)\in Y$. We want to show $z=y$, which reminds a lot of what injectivity can help proving -- so let us use the injective function we have by (ii), $g$: $$ g(z) = g( f\circ g(y) ) = g\circ f\circ g(y) = (g\circ f)( g(y ) ) = g(y) $$ the last equality by (i).

So $g(z)=g(y)$... but by (ii) we know $g$ is injective, which implies $z=y$.

$\endgroup$
  • $\begingroup$ woahhh nicee. I understand your proof pretty easy but still can't come up with solutions to these problems. How is the following true $I_x ∘ g(y)=g(y)$ $\endgroup$ – TheMathNoob May 26 '16 at 6:06
  • $\begingroup$ It is because g(y) is in X, and I_X takes as input an element in X and returns this same element. So I_X (g(y))=g(y) (for any y). $\endgroup$ – Clement C. May 26 '16 at 6:22
  • $\begingroup$ right.That's clear. I am new in number theory so I am just gonna keep working on it. $\endgroup$ – TheMathNoob May 26 '16 at 6:28
  • $\begingroup$ No worries -- and yes, I'd say the best approach towards pretty much anything in math is to practice a lot.. making mistakes helps also, quite a bit. $\endgroup$ – Clement C. May 26 '16 at 6:31
1
$\begingroup$

I think that for $g \circ f = I_X$ to be true, $g$ has to be injective.

Not necessarily. For a counterexample, take $X = Y = \mathbb N$ and define $f\colon \mathbb N \to \mathbb N$ and $g\colon \mathbb N \to \mathbb N$ by: $$ f(x) = 2x \qquad\text{and}\qquad g(y) = \begin{cases} y/2 &\text{if $y$ is even} \\ 42 &\text{otherwise} \end{cases} $$ Observe that for all $x \in \mathbb N$, we have that $g(f(x)) = g(2x) = x$ so that $g \circ f = I_\mathbb N$. But $g$ is not injective, since for example: $$ g(3) = 42 = g(5) $$


Show that if $g \circ f = I_X$ and $g$ is injective, then $f \circ g = I_Y$.

Given any $y \in Y$, we want to show that $f(g(y)) = y$. To this end, choose any $y \in Y$ and let $x = g(y)$. Then since $g \circ f = I_X$, we know that: $$ g(f(x)) = x $$ In other words, we know that $g(f(g(y))) = g(y)$. But since $g$ is injective, we conclude that $f(g(y)) = y$, as desired. $~~\blacksquare$

$\endgroup$
  • $\begingroup$ In your first counterexample, $$g\circ f(-1) = g((-1)^2) = g(1) = 1\neq -1$$ so $g\circ f\neq I_X$. A "correct" counterexample must have $f$ injective, and one can take say $f\colon \mathbb{R}\to (0,1)$ to be a bijection (and then $g$ can be arbitrary on $\mathbb{R}\setminus(0,1)$, even say constant, but $g\mid_{(0,1)}$ a bijection from $(0,1)\to\mathbb{R}$). $\endgroup$ – Clement C. May 26 '16 at 5:46
  • $\begingroup$ @ClementC. Whoops, you're right. I changed my counterexample. $\endgroup$ – Adriano May 26 '16 at 5:53
0
$\begingroup$

For $g\circ f$ to be the identity, it is required that $f$ be injective and $g$ be surjective (but not necessarily that $g$ be injective or $f$ surjective). Prove this first. Then note that $f\circ g$ being the identity requires $g$ to be injective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.