Linear map from the product of vector spaces to the addition

Question

If we add the vector spaces $U_1$,$U_{2}\leq V$, we have $U_{1}+U_{2}=\{u_1+u_2|u_1 \in U_1, u_2\in U_2\}$.

Now, I thought I understood what this meant, but then I looked at this set in action:

If we have the linear map $f:U_1 × U_2 → U_1 + U_2$ given by $f(u_1, u_2) = u_1 + u_2$, this has $ker (f) \cong U_1 ∩ U_2$.

I don't understand why this is the kernel. Don't we want some $u_1$ and $u_2$ such that $u_1 + u_2=0$?

Answer 1

Yes, $\ker f=\left\{(u_1,u_2)\in U_1\times U_2\,\middle|\,u_1+u_2=0\right\}$. But what is this set? If $u_1+u_2=0$, then $u_1=-u_2$ and therefore $u_1,u_2\in U_1\cap U_2$. And if $u\in U_1\cap U_2$, then $(u,-u)\in\ker f$. Can you use this to define an isomorphism between $\ker f$ and $U_1\cap U_2$?

Answer 2

Yes but choosing $u_1\in U_1$ and $u_2\in U_2$ such that $u_1+u_2=0$ we have $u_1=-u_2$ and by the definition of a vector space $u_1\in U_2$ and $u_2\in U_1$, thus an element in the kernel also lies in the intersection of the two spaces. Also an element lying in the intersection is such that the inverse lies in the intersection, again by basic properties of a vector space.

Answer 3

It depends on what the sign $\simeq$ means. Of course it is not an equality of sets, since there are easy counterexamples. If it is an isomorphism, then the equation $u_1+u_2=0$ shows that both $u_1\in U_2$ and $u_2\in U_1$.