Prove $(U+W)/U$ and $W/(U\cap W)$ are isomorphic

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Let $U$ and $W$ be subspaces of a vector space $V$; then $U$ is a subspace of $U+V$ and $U\cap W$ is a subspace of $W$. Prove that the quotient spaces $$(U+W)/U \ \ \text{and}\ \ W/(U\cap W)$$ are isomorphic.
Hint: A coset of $U$ in (U+W) has the form $w+U$, and $w_1+U \equiv w_2 +U$ iff $w_1-w_2\in U$. (equivalence classes or coset)

The following is my work:

We have to prove $\phi: (U+W)/U \xrightarrow[]{\phi}W/(U\cap W)$ is isomorphism.

Note: $\tilde{x} = x+U$ mans $\{v=x+u : u\in U\}$, which is a set.

  1. onto: For each $x\in W$, the coset of $(U\cap W)$ that contains $x$ is $\tilde{x}=x+(U\cap W)$.
    Since $x\in W$, so $\tilde{x}=w+(U\cap W)$. Since $(U+W)\subset U$, we prove onto.
  2. one-to-one: Suppose $\tilde{x}_1=w_1+U$ and $\tilde{x}_2=w_2+U$ with $w_1-w_2\notin U$ (i.e. two different coset) such that $\phi(\tilde{x}_1)\equiv \phi(\tilde{x}_2)$.
    Suppose $\phi(\tilde{x}_i) \equiv w+(U\cap W)$ where $i=1,2$.

Now I am stuck here.

  1. How to show $\phi^{-1}(\phi(\tilde{x}_i)) $ points to the same coset such that we reach a contradiction? My problem is I cannot quite understand what the map does here and how to map; therefore, I cannot continue it.
  2. How to show $\phi^{-1}$ is isomorphism? Since $(U\cap W)\subset U$, I have no idea how does "onto" hold.

How to fix them?

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On

It's much simpler using the homomorphism theorem. The map $$ \alpha\colon W\to (U+W)/U \qquad \alpha(x)=x+U $$ is a linear map. Its kernel is $U\cap W$ and therefore we get an isomorphism $$ (U+W)/U\cong W/\ker\alpha=W/(U\cap W) $$ by the main homomorphism theorem.

Without the homomorphism theorem, you can directly define $$ \beta\colon W/(U\cap W)\to U+W/U $$ by declaring $\beta(w+U\cap W)=w+U$

You have to check that

  1. $\beta$ is well defined: if $w_1+U\cap W=w_2+U\cap W$, then $w_1+U=w_2+U$

  2. $\beta$ is injective: if $w+U=0+U$ then $w+U\cap W=0+U\cap W$

  3. $\beta$ is surjective: this is the easy part, because $u+w+U=\beta(w+U\cap W)$