Show that f o g is one-to-one

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January 24th 2010, 04:54 PM
450081592

Show that f o g is one-to-one

(a) Show that the composition of two one-to-one functions, f and g, is one-to-one

(b) Express $(f o g)^{-1}$ in terms of $f^{-1} and g^{-1}$ .
January 24th 2010, 05:50 PM
Sudharaka

Dear 450081592,

1) Suppose a,b $\in{Dom (f_og)}$ such that,

$f_o{g(a)}=p$ and $f_o{g(b)}=p$

Then, f[g(a)]=f[g(b)]

g(a)=g(b) since f is an one to one function.

a=b since g is an one to one function.

Therefore $f_o{g(a)}=f_o{g(b)}\Rightarrow{a=b}$

Hence $f_o{g}$ is an one to one function.

2) Supose g(a) = c and f(c) = b,

Therefore, $f_og(a)=b$

Since $f_og$ is one to one it is invertible,

$(f_og)^{-1}(b)=a$ -----------A

Since f and g are one to one they are invertible.

$g^{-1}(c)=a$ and $f^{-1}(b)=c$

Therefore $g^{-1}[f^{-1}(b)]=a$ -------B

From A and B,

$g^{-1}_of^{-1}(b)=(f_og)^{-1}(b)$ $\forall$ $b\in Dom{(f_og)}$
January 24th 2010, 06:29 PM
Drexel28

Quote:

Originally Posted by 450081592

(a) Show that the composition of two one-to-one functions, f and g, is one-to-one

(b) Express $(f o g)^{-1}$ in terms of $f^{-1} and g^{-1}$ .

Quote:

Originally Posted by Sudharaka

Dear 450081592,

1) Suppose a,b $\in{Dom (f_og)}$ such that,

$f_o{g(a)}=p$ and $f_o{g(b)}=p$

Then, f[g(a)]=f[g(b)]

g(a)=g(b) since f is an one to one function.

a=b since g is an one to one function.

Therefore $f_o{g(a)}=f_o{g(b)}\Rightarrow{a=b}$

Hence $f_o{g}$ is an one to one function.

2) Supose g(a) = c and f(c) = b,

Therefore, $f_og(a)=b$

Since $f_og$ is one to one it is invertible,

$(f_og)^{-1}(b)=a$ -----------A

Since f and g are one to one they are invertible.

$g^{-1}(c)=a$ and $f^{-1}(b)=c$

Therefore $g^{-1}[f^{-1}(b)]=a$ -------B

From A and B,

$g^{-1}_of^{-1}(b)=(f_og)^{-1}(b)$ $\forall$ $b\in Dom{(f_og)}$

Jeez you two, haha. Try... \circ $f\circ g$

Also, for the second one there is a particularly nice result if your functions are both mappings from a set $X$ into itself. It follows that $f,g\in S_X$ (the permutation group on $X$ ) from where it follows from basic group theory that $\left(ab\right)^{-1}=b^{-1}a^{-1}$ .