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Show that if $M$ is a direct sum of $M_1$ and $M_2$ then $M/M_1$ is isomorphic to $M_2$ and $M/M_2$ is isomorphic to $M_1$.

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Well, if $M = M_1 \oplus M_2$, then every $m \in M$ can be written as $m=m_1+m_2$ where $m_1 \in M_1$ and $m_2 \in M_2$.

If you consider the linear map $\phi : M \to M_1$ which send every $m\in M$ to his projection over $M_1$, namely $m_1$, then the kernel of this map is exactly the set $M_2$, because $M_1\cap M_2 = \{0\}$.

However, this map is clearly onto, so by factorisation, $M/Ker\phi \cong Im\phi$, namely $M/M_2 \cong M_1.$

The other isomorphism can be deducted this way.

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You can simply view the direct sum as the set $M:=\{(x,y)\mid x\in M_1, y\in M_2\}$ and component-wise addition and scalar multiplication, together with the obvious embeddings $i_1\colon M_1\to M, x\mapsto (x,0)$ and $i_2\colon M_2\to M, y\mapsto (0,y)$. In this setup, it is clear that $M/i_2(M_2)\to M_1, (x,y)+i_2(M_2)\mapsto x$ and $M/i_1(M_1)\to M_2, (x,y)+i_1(M_1)\mapsto y$ are isomorphisms.

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I like to prove such things only using the universal properties..

Firstly a map out of the coproduct $M=M_1 \oplus M_2$ is the same as maps out of each guy individually. Moreover by the universal property of quotients maps out of $M/M_1$ are maps out of $M$ that vanish on $M_1$, so you directly see that $M_2$ is a quotient (recall that objects defined by a universal property, such as coproducts and quotients are best thought of as unique only up to unique isomorphism).

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