# Surjections and a true/false statement.

• Dec 15th 2008, 03:50 AM
Showcase_22
Surjections and a true/false statement.
This is a question I have on a test paper:

Quote:

If $\displaystyle f:A \rightarrow B$ and $\displaystyle g:C \rightarrow D$ are surjections where $\displaystyle Im(f) \subset C$, then g o f: $\displaystyle A\rightarrow D$ 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.
• Dec 15th 2008, 07:24 AM
flyingsquirrel
Hi,

Take a look at this diagram :

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

Edit: That's my 666th post. :D
• Dec 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 $\displaystyle B \subset C$ 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.
• Dec 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 $\displaystyle B \subset C$ there will always be a point that is not connected to a point in a.

Therefore it cannot be a surjection.

That's it!