Composition of Maps

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August 23rd 2009, 08:20 PM
Roam

Composition of Maps

Let $A,B$ be sets and let $f:A \rightarrow B$ and $g,h:B \rightarrow A$ be functions.

(1) Suppose $h o f$ is an injective map from $A$ to itself. Show that $f$ is injective.

(2) Suppose now that $f o g = 1_{B}$ and $hof = 1_{A}$ . Show that $f$ is bijective and $g=h$ .

My attempt:

(1) $h o f$ is 1-1 $\Leftrightarrow \forall a,a' \in A$ such that $a \neq a'$ , $h o f(a) \neq h o f(a')$

$\Leftrightarrow h(f(a)) \neq h(f(a'))$

$\Leftrightarrow f(a) \neq f(a')$

It follows that $a \neq a'$ . Is this proof correct? Do I need to add any more explanation?

(2) I need some help with question two, so I can get started. Btw, the $1_A$ and $1_B$ notation represents the identity.

Thanks in advance,
August 24th 2009, 12:02 AM
Liwuinan

Proof of (1) is basically correct except for some typos:

Quote:

Originally Posted by Roam

It follows that $a \neq a'$

This is where you began, what you proved is that $f(a) \neq f(a')$ . Also, $h(f(a)) \neq h(f(a'))$ is not equivalent to
$f(a) \neq f(a')$ , but the implication from left to right holds, and that is what you need.

(2) $f$ is injective: apply (1) to $h \circ f = 1_A$ is injective.
$f$ is surjective: if it is not, then $f \circ g$ also isn't surjective, but $f \circ g = 1_B$ is surjective.
Further, we have $g= 1_A \circ g=h \circ f \circ g=h \circ 1_B=h$ , because composition of functions is associative.

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