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?

surjective: Not sure how to go about this... Any advice? Thanks

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- Oct 8th 2010, 11:04 PMjzelltShow 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?

surjective: Not sure how to go about this... Any advice? Thanks - Oct 9th 2010, 02:41 AMSwlabr
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 $\displaystyle f(a, b) = x$? - Oct 9th 2010, 03:28 AMHallsofIvy
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.

Quote:

Is this the correct approach?

surjective: Not sure how to go about this... Any advice? Thanks

If you are given a number b does there exist a pair (x, y) such that f(x,y)= b? If so, what is it?