Getting an isomorphism from a short exact sequence of inner product spaces Let $L,M,N$ be finite dimensional inner product spaces and $0 \to L \xrightarrow{\alpha} M \xrightarrow{\beta} N \to 0$ is a short exact sequence. Now let $\beta^* : N \to M $ be the adjoint map (the dual space has been identified with the original space now).

Is it true that :  $\alpha \oplus \beta^* : L \oplus N \to M$ is an isomorphism ?

If so, how to prove it ? 
 A: Note that $\beta^*:N\to M$ is the unique linear map satisfying
$$
\langle\beta(m),n\rangle_M = \langle m,\beta^*(n)\rangle_M
$$
The following are general facts about adjoints.

Fact 1. $\DeclareMathOperator{image}{image}\image(\beta^*)=\ker(\beta)^\perp$
Fact 2. $\ker(\beta^*)=\image(\beta)^\perp$

Can you prove these facts?
These two facts together with the exactness of
$$
0\to L\xrightarrow{\alpha}M\xrightarrow{\beta}N\to0
$$
imply that $\image(\beta^*)=\image(\alpha)^\perp$ and $\ker(\beta^*)=0$.
Now, your map $\alpha\oplus\beta^*: L\oplus N\to M$ is
$$
(\alpha\oplus\beta^*)(l,n)=\alpha(l)+\beta^*(n)
$$
To see this is injective, suppose $(\alpha\oplus\beta^*)(l,n)= 0$. Then
$$
\alpha(l)=\beta^*(-n)
$$
It follows that $\alpha(l)\in\image(\alpha)$ and $\alpha(l)\in\image(\beta^*)=\image(\alpha)^\perp$. Since
$$
U\cap U^\perp=0
$$
for any subspace, it follows that $\alpha(l)=0$. Since $\alpha$ is injective we conclude $l=0$.
Finally, we see 
$$
\beta^*(-n)=\alpha(l)=0
$$
Since $\beta^*$ is injective, it follows that $-n=0$ whence $n=0$.
Hence $(l,n)=(0,0)=0$ and $\alpha\oplus\beta^*$ is injective.
