If restriction and quotient of a linear operator is nonsingular, the operator is itself nonsingular.

Question

Let $V$ be a vector space over a field $\mathbb{F}$. $V$ is not necessarily of finite dimensional and let $T$ be a linear operator on $V$. Let $W$ be a $T$-invariant subspace of $V$. Consider the induced operator $\left.T\right|_W$ on $W$ and $\bar{T}$ on $V/W$.We have to show the following:

If $\left.T\right|_W$ and $\bar{T}$ both are isomorphism then so is $T$.

I have shown that $T$ is injective. To see this, let $x\in \ker{T}$. Then $\bar{T}(\bar{x})=0$ and since $\bar{T}$ is injective, we obtain $x \in W$. Now $\left.T\right|_W(x)=0$ implies that $x=0$, i.e., $T$ is injective. In order to show that $T$ is surjective consider $y\in V$. Since $\bar{T}$ is surjective, $\bar{T}(\bar{z})=\bar{y}$ for some $z \in V$. Then $z-y \in W$.

How can I continue from here to complete it?

Need some help. Many thanks.

Answer 1

You can conclude, from $\overline{T}(\overline{z}) = \overline{y}$ that $T(z) - y \in W$, not $z - y \in W$. Or, even better, $y - Tz \in W$.

From the fact that $T|_W$ is surjective, some $w \in W$ exists such that $$y - Tz = Tw \implies y = T(z + w),$$ which shows $T$ is surjective.