Disprove the statement: If $g\circ f=I_X$then $f\circ g=I_Y$.

Question

If someone could walk me through this I would greatly appreciate it.

Disprove the following statement:

Let $f : X \rightarrow Y$ and $g : Y \rightarrow X$ be functions. If $g\circ f=I_X$then $f\circ g=I_Y$

Answer 1

The first counterexample one tries should be the simplest. Let $$X=\{\#\}, \ \ Y = \{*,\&\}.$$ Define $$f(\#) = * \ \mbox{ and } \ g(*) = g(\&) = \# $$ Note that $$ (g\circ f)(\#) = \#$$ Now examine $(f\circ g)$. Do you see the issue?

Answer 2

Let $f$ be the natural inclusion of the real line into the plane, and let $g$ be the natural projection of the plane onto the real line.

Answer 3

Let $X=\{0\}$ and $Y=\mathbb{Z}$. Now define $f:X\to Y$ by $f(0)=0$ and $g:Y\to X$ by $g(x)=0$.

Now $g\circ f= I_X$ and $f\circ g\neq I_Y$ since $f\circ g(1)=f(0)=0\neq I_Y(1)$.

Answer 4

Let the domain of $f$ be the integers and the domain of $g$ be the reals. Then take

$$f(n) = n \\ g(x) = \lfloor x \rfloor$$

It is easy for see that for all $n$ in the domain of $f$, $$g(f(n)) = \lfloor n \rfloor = n$$ so $g\circ f = I_X$. Yet for real $x$ (in the domain of $g$) $$ f(g(x)) = \lfloor x \rfloor \neq x $$ unless $x$ happens to be a integer. So $f\circ g \neq I_Y$

Now try the same problem but specifying that $X$ and $Y$ be the same space. You can still find a counterexample. (Hint - the drange of $X$ need not be all of $Y$).