1. ## 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.

2. Consider $A = \left\{ {1,2,3} \right\}\;,\;B = \left\{ {a,b,c,d} \right\}\;\& \;C = \left\{ {x,y} \right\}$.
Define functions: $f:\left\{ {(1,a),(2,b),(3,c)} \right\}\;\& \;g:\left\{ {(a,x),(b,x),(c,y),(d,y)} \right\}$.

3. Originally Posted by Plato
Consider $A = \left\{ {1,2,3} \right\}\;,\;B = \left\{ {a,b,c,d} \right\}\;\& \;C = \left\{ {x,y} \right\}$.
Define functions: $f:\left\{ {(1,a),(2,b),(3,c)} \right\}\;\& \;g:\left\{ {(a,x),(b,x),(c,y),(d,y)} \right\}$.
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..

4. 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.

5. ## Injective and Surjective and Bijective (- oh my!)

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

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

7. 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, $\phi: \mathbb{N} \rightarrow \mathbb{N}$, $x \mapsto x^2$. Then, taking the constant map, $\theta: \mathbb{N} \rightarrow \mathbb{N}$, $x \mapsto x$.

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