i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity?

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- May 31st 2010, 01:49 AMalexandrabel90kernel
i was reading my notes and it states that the kernel of a function allows you to check if the function is injective.

how do you check with a kernel? isint elements in the kernel, elements that through the function get mapped to the identity? - May 31st 2010, 01:59 AMSwlabr
- May 31st 2010, 02:11 AMalexandrabel90
im studying group homomorphisms now.

in this case, if it is a function from A to B, where the identity of B =0, then the element in A has to be 0 since u=0 (from your above statement) right?

what if the element in A that get mapped to 0 in B is not 0, wont u be not equals to 0? in such case, Tu=0 but u wont be zero?

thanks! - May 31st 2010, 02:26 AMSwlabr
My above proof can be quite easily changed to look at group homomorphisms (just change a+b to ab, and a-b to ). So I'll use group notation.

Is your question basically,

Is it possible that ?

The answer is `no'.

Let . Then . Thus, ( is an idempotent). Then, . Thus, for a group homomorphism.

The same holds for linear maps, etc. - May 31st 2010, 02:38 AMalexandrabel90
for the proof under group homomorphism, we are assuming that it is under the operation x right? is that what we usually assume?

becos im not sure when it should be under + and when it should be under x.

thanks - May 31st 2010, 02:42 AMSwlabr
- May 31st 2010, 05:19 AMHallsofIvy