Getting an isomorphism from a short exact sequence of inner product spaces

Question

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 ?

Answer 1

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.