Surjections and a true/false statement.

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December 15th 2008, 03:50 AM
Showcase_22

Surjections and a true/false statement.

This is a question I have on a test paper:

Quote:

Is the following statement always true? Justify your answer.

If $f:A \rightarrow B$ and $g:C \rightarrow D$ are surjections where $Im(f) \subset C$ , then g o f: $A\rightarrow D$ is a surjection.

(I wrote "g o f" when I mean the "composite function of g and f").

I would say this statement is true because the domain of g is contained in the image of f (which is a necessary criterion for g o f to exist).

The composite function of two surjections is also a surjection.

Both these facts mean that the statement is always true.

I think this is right, I just wanted to see if I had overlooked anything.
December 15th 2008, 07:24 AM
flyingsquirrel

1 Attachment(s)

Hi,

Take a look at this diagram :

Attachment 9265

$f$ maps $A$ onto $B$ , $g$ maps $C$ onto $D$ and $\mathrm{Im}(f)=B\subset C$ so, according to you, $g\circ f$ should map $A$ onto $D$ . Is that true ?

Edit: That's my 666th post. :D
December 17th 2008, 03:35 AM
Showcase_22

Right, i'd say that the statement is false then.

The third point in the set C (the point that is in C but not in B) isn't connected to a point in A and since $B \subset C$ there will always be a point that is not connected to a point in a.

Therefore it cannot be a surjection.

I think i've got it.
December 17th 2008, 04:43 AM
flyingsquirrel

Quote:

Originally Posted by Showcase_22

The third point in the set C (the point that is in C but not in B) isn't connected to a point in A and since $B \subset C$ there will always be a point that is not connected to a point in a.

Therefore it cannot be a surjection.

That's it!