# Thread: subspace and isomorphism

1. ## subspace and isomorphism

Let U be a vector space, U x U the product vector space, and

# = { (u,u)|u in U} ⊆ U x U

Prove that # is a subspace of U x U and that (U x U) / # (isomorphism) U.
plx help....

2. Originally Posted by mathbeginner
Let U be a vector space, U x U the product vector space, and

# = { (u,u)|u in U} ⊆ U U.

Prove that # is a subspace of U x U and that (U x U) / # (isomorphism) U.

plx help....
What about the obvious choice of $T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" alt="T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" />. This is evidently surjective since $T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}$ and it's a homomorphism since $T(\alpha(u_1,u_2)+\beta(u_3,u_4))=\left(\alpha u_1+\beta u_3,\alpha u_2+\beta u_4)=(\alpha u_1+\beta u_3)-(\alpha u_2+\beta u_4)=\alpha(u_1-u_2)+\beta(u_3-u_4)=\alpha T((u_1,u_2))+\beta T((u)3,u_4))$. I claim that we are now finished. Why?

3. Originally Posted by Drexel28
What about the obvious choice of $T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" alt="T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" />. This is evidently surjective since $T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}$ and it's a homomorphism since $T(\alpha(u_1,u_2)+\beta(u_3,u_4))=\left(\alpha u_1+\beta u_3,\alpha u_2+\beta u_4)=(\alpha u_1+\beta u_3)-(\alpha u_2+\beta u_4)=\alpha(u_1-u_2)+\beta(u_3-u_4)=\alpha T((u_1,u_2))+\beta T((u)3,u_4))$. I claim that we are now finished. Why?
why we only prove that # is a subspace of U then it is homomorphism
and I don't reli get why This is evidently surjective since $T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}$

4. Originally Posted by mathbeginner
why we only prove that # is a subspace of U then it is homomorphism
and I don't reli get why This is evidently surjective since $T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}$
Let $u\in U$ then [math}T(u,0)=u-0=u[/tex]