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### Functions and Inverses

```Date: 05/29/2002 at 01:42:57
From: Marianne Richa
Subject: Functions (Surjection, Bijection)

Hi Dr. Math,

I was doing the following exercise and I couldn't find when both are
injective and surjective:

Let A, B, and C be three sets, and f:A->B, g:B->C are two
functions.

Define A, B, C, f and g such that (g o f) is both injective and
surjective, but f is not surjective and g is not injective.

Best Regards
Marianne
```

```
Date: 05/29/2002 at 12:53:52
From: Doctor Peterson
Subject: Re: Functions (Surjection, Bijection)

Hi, Marianne.

Let's try to picture it. Since f is not surjective, the image of A
(range of f) must be only part of B:

+---------------+
|       B       |
+-----+           |    +-----+    |
|     |      f    |    |     |    |
|  A  |--------------->|  A' |    |
|     |           |    |     |    |
+-----+           |    +-----+    |
|               |
+---------------+

But when we follow that with g, the composition must be both
injective and surjective, so C must be "the same size as" A, and must
be the image of A':

+---------------+
|       B       |
+-----+           |    +-----+    |             +-----+
|     |      f    |    |     |    |     g       |     |
|  A  |--------------->|  A' |----------------->|  C  |
|     |           |    |     |    |             |     |
+-----+           |    +-----+    |             +-----+
|               |
+---------------+

Now g is defined on all of B, so clearly it must not be injective; it
must take the rest of B into C as well.

There are many ways to do this. One easy way is to take A, B, and C
as the integers, and take f as

f(x) = 2x

This is not surjective, but is injective; its image is the even
integers. Now can you find a g that will take these to all integers?

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/29/2002 at 14:23:51
From: Marianne Richa
Subject: (Continued)

Hi,

I understood all that you wrote, but still I couldn't define
g(x).  Could it be 3x?

Thank you.
```

```
Date: 05/29/2002 at 14:41:59
From: Doctor Peterson
Subject: Re: (Continued)

Hi, Marianne.

You don't sound very sure ... so you probably aren't right! Did
you try finding the composite (f o g) and seeing if it fits the
requirements; or check whether your g is surjective?

You want a function g that is sort of an inverse of f; for every
integer, there should be some even number that g will take there.
Since we made f double each integer, what if g divides by 2? The
problem is, the odd integers won't go to integers then, so you'll
need a different definition for the function in that case. If you
think about it, you'll realize that it really doesn't matter what
you choose to do with the odd numbers (which are B-A' in my picture).

You might like to try a different solution. Suppose we just made
A = {1,2,3} and B = {4,5,6,7} and C = {8,9,10}. Can you follow my
picture to define f and g by just naming where each function takes
each element of its domain?

The concepts of injective and surjective functions become almost
trivial if you learn to picture them right. This page may help a
little:

http://mathforum.org/library/drmath/view/52454.html

You can find more help by searching our site for those words.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Functions
High School Sets

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