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
What are we "in" here? Linear maps? Group homomorphisms?

The proof is pretty much the same in every context through.

Let $\displaystyle T$ be a linear map.

Then $\displaystyle T$ is injective

$\displaystyle \Leftrightarrow (Tx = Ty \Rightarrow x=y) $

$\displaystyle \Leftrightarrow (Tx-Ty = 0 \Rightarrow x=y) $

$\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x = y) $

$\displaystyle \Leftrightarrow (T(x-y) = 0 \Rightarrow x-y = 0) $

$\displaystyle \Leftrightarrow (Tu = 0 \Rightarrow u =0) $

$\displaystyle \Leftrightarrow kerT = \{0\}$. - 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 $\displaystyle ab^{-1}$). So I'll use group notation.

Is your question basically,

Is it possible that $\displaystyle 1 \not\in ket(\phi)$?

The answer is `no'.

Let $\displaystyle 1 \phi = a$. Then $\displaystyle a^2 = (1\phi)^2 = (1^2)\phi = 1\phi = a$. Thus, $\displaystyle a^2=a$ ($\displaystyle a$ is an idempotent). Then, $\displaystyle a^2 = a \Rightarrow a^2a^{-1} = aa^{-1} \Rightarrow a=1$. Thus, $\displaystyle 1_G\phi = 1_H$ for $\displaystyle \phi: G\rightarrow H$ 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
Just go through the proof I gave for linear maps and replace $\displaystyle +$ with $\displaystyle *$ and $\displaystyle T$ with $\displaystyle \phi$.

So it starts,

$\displaystyle \phi$ is injective

$\displaystyle \Leftrightarrow (g\phi = h\phi \Rightarrow g=h)$

$\displaystyle \Leftrightarrow ((g\phi)(h\phi)^{-1} = 1 \Rightarrow g=h)$

$\displaystyle \vdots$ - May 31st 2010, 05:19 AMHallsofIvy