Thread: Surjective, Bijective

Prove or disprove the following statement: if f is an injective function from A to B, and g is a surjective function from B to C, then gof is a bijective function from A to C.

Really need help. I'm so lost.

Consider .
Define functions: .
Use those to think about this problem.

Originally Posted by Plato

Consider .
Define functions: .
Use those to think about this problem.

but according to this example g0f(1) = x and g0f(2) = x so this violates the definition on a one-one function. so according to this example the function g0f is not 1-1..

Originally Posted by doresa

but according to this example g0f(1) = x and g0f(2) = x so this violates the definition on a one-one function. so according to this example the function g0f is not 1-1..

That's the point. It said prove or disprove.

You might find Chapter 3 of Part I of Topology and the Language of Mathematics useful. It provides a number of example of problems like yours, with solutions.

A free download of the first 50 pages (which includes Chapter 3 of Part I) is available here:
Bobo Strategy - Topology

Hope it's useful -
Chris

its somewhat usefull. but its more on the topology..

Originally Posted by extremely_lost

Prove or disprove the following statement: if f is an injective function from A to B, and g is a surjective function from B to C, then gof is a bijective function from A to C.

Really need help. I'm so lost.

Making the second of your maps the constant map while the first is a non-surjective injection will provide quite a neat solution.

For instance, , . Then, taking the constant map, , .

Here, the composition of and is just the first map, \phi. Clearly it is injective but not surjective.