Proof of Proposition

**shilz222** · August 14th 2007, 09:58 PM

How do you prove the following? Let $f: A \to B$ .

(1) $f$ is injective if it has a left inverse.
(2) $f$ is surjective if it has a right inverse.
(3) $f$ is bijective if it has both left and right inverse.

(4) if $|A| = |B|$ then $f: A \to B$ is bijective if and only if $f$ is injective if and only if $f$ is surjective.

Thanks

**Plato** · August 15th 2007, 03:37 AM

Originally Posted by shilz222

(1) $f$ is injective if it has a left inverse.
(2) $f$ is surjective if it has a right inverse.

You must give us detailed definitions for left and right inverses of a function.
Having taught this material from at least ten different texts, I have never seen those words used in the context.
BTW: what textbook are you using for your study?

You should know that #3 is the usual definition for bijectivity.

Of course, #4 follows at once from the others.

**shilz222** · August 15th 2007, 08:59 AM

A left inverse is a function $g: B \to A$ such that $g \circ f: A \to A$ is the identity map on $A$ .

A right inverse is a function $h: B \to A$ such that $f \circ h: B \to B$ is the identity map on $B$ .

Using Abstract Algebra by Dummit and Foote.

This is what I think: The inverse function returns elements in the domain (i.e. inverse image). So for the left inverse you get back the same value in the domain. For the right inverse you get back the same value in the codomain.

**Plato** · August 15th 2007, 10:10 AM

Before I go into this a word of warning. I have the second edition of Dummit & Foote. It is known as a difficult text and it uses non-standard notation in some cases.

If $f:A \mapsto B$ ,
i) Suppose $g$ is a left inverse of $f$ . If $f(x)=f(y)$ then $g \circ f(x)= g \circ f(y)$ but by definition $g \circ f(x)= x$ and $g \circ f(y)= y$ therefore $x=y$ . That means that $f$ is injective.

ii) Suppose that $f$ is injective. We know that $f\left( A \right) \subseteq B$ is the image set. Because $f$ is injective then for each $p \in f\left( A \right)$ there is exactly one $a_p \in A$ such that $f\left( {a_p } \right) = p$ . Now we define $<br /> g:B \mapsto A$ . But first we must fix a point $q \in A$ .
$g(t) = \left\{ {\begin{array}{lr}<br /> {a_t ,} & {t \in f(A)} \\<br /> {q,} & {t \in \left( {B - f(A)} \right)} \\<br /> \end{array}} \right.$
It is clear that $g$ is a left inverse.

**shilz222** · August 15th 2007, 10:16 AM

Is is a good text to self study from? What would you recommend?

**Plato** · August 15th 2007, 10:30 AM

Here is a most readable text.
Amazon.com: Introduction to Advanced Mathematics (2nd Edition): Books: William J Barnier,Norman Feldman
It is self-contained, easy to follow, has all the information you have been asking about.
But best of all it uses standard notation in almost every case.

**shilz222** · August 15th 2007, 10:40 AM

I have a book like that. I am studying multiple mathematical topics at the same time (abstract algebra, analysis, and transition course currently) (alternating between them).

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