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

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.

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.