$f \circ g =\operatorname{ id}$ and $g \circ f \neq \operatorname{id}$?

Question

Are there two functions $f$ and $g$ s.t. $$f \circ g = \operatorname{id}$$ but $$g \circ f \neq \operatorname{id}?$$ Could someone give an example or a proof that this is impossible?

This must be trivial, but I can't figure it out. :) Thanks!

Answer 1

HINT: Consider $\sin$ and $\arcsin$ on their natural domains.

Answer 2

It is possible to define the functions via matrices, in which case the product of the matrices is not commutative - I don't know if that is quite what you are looking for. You could also define the functions by $f(x) = x^2$, $g(x) = \sqrt{x}$, where $f:\mathbb{R} \to [0 , \infty)$ and $g: [0 , \infty) \to [0 , \infty)$. Here, $f(g(x)) = x $ but $g(f(x)) = |x|$. There are of course other examples of similar type.

Answer 3

Hint: Consider the two functions on the integers: $$ f(x)=\left\lfloor\frac x2\right\rfloor $$ and $$ g(x)=2x $$

Answer 4

Let $A=\{0,1\}$ and $B=\{0\}$ and define the functions $f:A\to B$ and $g:B\to A$ so that $f(0)=f(1)=0$ and $g(0)=0$. Then $f\circ g:B\to B$ is the identity but $g\circ f:A\to A$ maps everything to $0$.

Answer 5

Let $f:A\rightarrow B$ and $g:B\rightarrow A$.

If $B$ is a singleton then $f\circ g=\operatorname{id}_B$.

($\operatorname{id}_B$ is unique as function $B\rightarrow B$)

If $B$ is a singleton and $A$ is not a singleton then $g\circ f\neq \operatorname{id}_A$.

($g\circ f$ is constant and $\operatorname{id}_A$ is not)

---

Nice to remember:

$s\circ i=\operatorname{id}$

Here $s$ stands for surjective and $i$ for injective. So from $f\circ g=\operatorname{id}$ you are allowed to conclude that $f$ is surjective and $g$ is injective. A more general version (for later, if there is some familiarity with categories) is:

$r\circ s=\operatorname{id}$

Here $r$ for retraction and $s$ for section.

These mnemonics are very valuable to me.

Answer 6

To treat this matter properly, one must consider the functions as having well-defined domains and co-domains (or sources and targets, if preferred). For the above compositions to even make sense, we must have $$ f: A \to B $$ and
$$ g: B \to A . $$ Now the compositions are $$ f\circ g: B \to B $$ and
$$ g\circ f: A \to A, $$ so if $A$ and $B$ are different, the two compositions have no chance of being the same.

BUT, even if $f\circ g = id_B$, we still need not get $g\circ f = id_A$.

Example: $A = \{ -1 , 1 \}$ and $B = \{ 1 \}$ with $g(1) = 1$ and $f(-1) = f(1) = 1$.

(This is just the square/squareroot example that was mentioned earlier.)

Bill Taylor

Answer 7

Take the space of polynomials, $f = $ differentiation and $g = $ integration.

Then $f \circ g = \operatorname{id}$ but $g \circ f \ne \operatorname{id}$ because $g(f(x^2+1))=g(2x)=x^2$.

Answer 8

To work out a pair of function $f$ and $g$ that satisfies these conditions the following lemmas are helpful.

Lemma 1:

If $$f \circ g = \operatorname{id} \tag{1}$$ then $f$ is surjective and $g$ is injective.

$\blacksquare$

I ommit the proof of this well known lemma. From lemma 1 follows the next lemma.

Lemma 2:

If $$f \circ g = \operatorname{id}$$ and $g$ is surjective then $$g \circ f = \operatorname{id}$$

$\blacksquare$

Proof:

$g$ is injective by lemma 1 and surjective, so it is bijective and $f$ is its inverse.

$\blacksquare$

So let's assume that $$g:A \mapsto B$$ satisfies $(1)$. So $g$ must be a injective but not surjective. We define $$C:=g(B)$$ Then $$C \subsetneqq B$$ and $$g:A \mapsto C$$ is bijective. We define $$f|_C: C \mapsto A$$ as $$f|_C:=g^{-1}|_C$$ and define $$f|_{B\setminus C}: B\setminus C \mapsto A $$ arbitrary. The $f$ and $g$ have the desired properties:

$$f \circ g = \operatorname{id} $$ because $$f|_C \circ g = \operatorname{id}$$ We have $$(g \circ f)|_C=\operatorname{id}|_C$$ But for $b \in B \setminus C$ we have $$(g \circ f) (b)=c \in C$$ and so $b \ne c$. But we have $$(g \circ f) (b) \ne c$$ and so $g \circ f$ is not injective.

Example 1

Choose an injective but not surjective function $g$, e.g. $$\exp: \to e^x$$ $$\exp: \mathbb{R} \mapsto \mathbb{R}$$ Then we have $$A=\mathbb{R}$$ $$B=\mathbb{R}$$ $$C=\mathbb{R^+}$$ We have $$f|_\mathbb{R^+}=\log$$ and we define $$f: x \to x, \; \forall x \in \mathbb{R_0^\textbf{-}} $$

$\blacksquare$

Example 2

$M$ is an arbitrary set with $m \in M$ and $q \notin M$. We define $$A:=M$$ $$B:=M \cup \{q\}$$ and the functions $g$ $$g:M \mapsto M \cup \{q\}$$ $$g: x \to x$$

We have $$C=M$$ and $$f:M \cup \{q\} \mapsto M $$ $$f: x \to \begin{cases} x, \; \forall x \in M \\ m , \mbox{if } x=q \end{cases} $$

$\blacksquare$