1. ## Show Injective/surjective

f: A x b --> A defined by f(x,y) = x

Injective: Here's what I did
Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
Is this the correct approach?

2. For injective, you need to say that $\displaystyle b$ is not equal to $\displaystyle y$ (you start with $\displaystyle (a, b) \neq (x, y)$, but then you concluded that $\displaystyle a=x$ but not that $\displaystyle b=y$).

So, a proof would be: Note that for $\displaystyle b \neq y$, $\displaystyle f(x, b) = f(x, y)$ but $\displaystyle (x, b) \neq (x, y)$.

What you did was write down the working to find the proof, but it wasn't the proof itself per se.

For surjective, you start with something arbitrary in the image, A, and find something which maps to it. So, let $\displaystyle x \in A$. Can you think of some $\displaystyle (a, b) \in A \times B$ such that $\displaystyle f(a, b) = x$?

3. Originally Posted by jzellt
f: A x b --> A defined by f(x,y) = x

Injective: Here's what I did
Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
Well, what causes the "thus"? To show "not injective", you need to assert that two different pairs, $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$, can give the same value- you need to give a counterexample.

Is this the correct approach?