Functions on Arbitrary Sets and Groups

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

For dummies like me, that example helped me out GREATLY! i was finally able to visualize this problem.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

looking back now for #3, i initially said it wasn't injective, but every element in A would have a unique image in AxB wouldn't it?

i could show?
if f(x)=(x,b) & f(y)=(y,b)
then we see x=y
so (x,b)=(y,b)

daveyinaz · **Joined:** Sat, 16 Aug 2008 04:47:19 UTC **Posts:** 208

You are right about it being injective, but you have to follow the definition of injectivity and that is "if f(x) = f(y)

, then

.

so in this case you must start with f(a) = f(a')

and end with a = a'

for some $a, a' \in A$ .

Really the hard part is putting into words what comes out in the middle...

Shadow · **Posted:** Thu, 1 Mar 2012 02:49:17 UTC

daveyinaz wrote:

You are right about it being injective, but you have to follow the definition of injectivity and that is "if f(x) = f(y)

, then

.

so in this case you must start with f(a) = f(a')

and end with a = a'

for some $a, a' \in A$ .

Really the hard part is putting into words what comes out in the middle...

I disagree, the definition is definitely one way to do this, but you can also easily produce a left-inverse as well.

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Functions on Arbitrary Sets and Groups

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