Second isomorphism theorem for subspaces

Question

just like I did some days ago, I now have to show that

$T/T\cap U \cong (U+T)/U $.

Therefore I tried finding a surjective homomorphism and then, by using the first isomorphism theorem, I should be able to show this.

So I took a look at: $ h: T \to (U+T)/U $ with $ t \to t+U$.

1) it's surjective. 2) it should be a homomorphism. However, I'm not able to show this..

Answer 1

I prefer not to use the actual cosets like $t + U$. I will write $\overline{t}$ for this.

To show that $h$ is a homomorphism, consider it as the composite of two mappings that are obviously homomorphisms: the inclusion $i \colon T \to U + T$, and the projection $p \colon U + T \to (U + T)/U$.

To show that $h$ is surjective, Let $x \in (U + T)/U$, and we will show that $x$ is in the image of $h$. $x = \overline{y}$ for some element $y \in U + T$. Therefore $x = \overline{u + t}$ for some $u \in U, t \in T$. But $u + t \equiv t \ (\textrm{mod} U)$, so $x = \overline{u + t} = \overline{t} = h(t).$

Answer 2

Step I. Define the homomorphism $$ \varphi : T\to (U+T)/ U, $$ by $\varphi(t)=t+U$.

Step II. Show that $\varphi$ is onto. An arbitrary element of $(U+T)/U$ is of the form $u+t+U$, for some $u\in U$ and $t\in T$. But $u+t+U=t+U=\varphi(t)$. Thus $\varphi$ is onto.

Step III. Show that the kernel of $\varphi$ is $U\cap T$. Let $t\in T$, such that $\varphi(t)=0$ or $t+U=0$. But this is true if and only if $t\in U$. Hence $t\in T\cap U$. Inversely, if $t\in U\cap T$, then $t\in U$ and thus $\varphi(t)=t+U=0+U$.

Step IV. Apply the 1st isomorphism Theorem.