S.O.S. Mathematics CyberBoard

Im doing numbers 1,2,3. Im confused on the notation.

For #1.) and #2.) To prove its injective or one-to-one, what do i need to show?

for example in 1.) to show its injective i wanted to show f(a,b)=f(c,d) which then a=c. But this is wrong.
Should it be , suppose f(a,y)=f(c,y)?

The first function is not 1-1 unless B only has one element, so you cannot prove it is injective because it's more or less guaranteed NOT to be.

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... or: if A or B is empty.

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Ah yes, this pathology with the universal quantifier.

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So for #2). to prove its injective could i say if suppose f(a,b)=f(b,a) then (b,a)=(a,b)?

And #3 i'm at a loss as show if it's a bijection

#3 cannot be a bijection. Consider such that , then but ....assuming the cardinality of is greater than 1.

And for #2 to show something is injective, you can usually follow the definition.
If , then which means and , so .

For #2 just find a left-inverse and that shows your function is injective.

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Got it for #2.)

For #3) ito show its not injective i made f(a)=f(c), with f(a)=(a,b) & f(c)=(c,b). since (a,b)=(c,b) it shows b is repeated so its not 1-1 . But my teacher said i need to check whether a=c?

I really think you should just follow the definition of injectivity. Which means do #3 similar to way I told you about number #2 because ultimately you want to show, by definition, that if , then which is what your teacher is telling you.

For surjective. z in AxB, then z=(z,b)=f(z). So every f(x) for f(x)=(a,b)?

Maybe im having trouble visualizing what exactly. A->AxB looks like

Huh? Surely . In any case, it should be dead obvious that the map from cannot be surjective unless B has only one element.

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I don't know if you are having problems visualizing what looks like since it's explicitly given to you by the mapping where and is always the same element.

I'll give an example to see if it helps.
Let and and assume that , then .

Even with this example you should see why is not surjective with these particular sets and .

I think im starting to get it. So for example AxB contains for instance (2,8), which is not in the function f, so since it's not the image of anything then f is not surjective

Yes. That typically is how to show something is not surjective. Find one element in the range that is not hit by any element in the domain or as you put it...not in the image.

Of course, I put this more mathematically in my first post.