Show that the statements about maps are equivalent

Question

Let $\mathbb{K}$ be a field and $V,W,X$ $\mathbb{K}$-vector spaces.

Let $\phi:V\rightarrow W$, $\psi:W\rightarrow X$ be linear maps and let $\phi$ be injective and let $\psi$ be surjective.

I want to show that the following statements are equivalent:

(i) $\psi\circ \phi$ is an isomorphism of vector spaces.

(ii) It holds that $W=\text{Image}(\phi)+\ker (\psi)$ and $\text{Image}(\phi)\cap \ker (\psi)=\{0\}$.

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For the direction $(i)\rightarrow (ii)$ do we use the rank nullity theorem?

Answer 1

It's easy to show that if (i) holds, then $\mathrm{im}\,\phi \cap \ker\psi = 0$. Obviously $\mathrm{im}\,\phi + \ker\psi$ is a subspace of $W$, so we only need to show that the dimensions are the same: $$\mathrm{rk}\,\phi + \mathrm{null}\, \psi = \dim W.$$ You mentioned the rank-nullity theorem, which is a perfect tool for this. (Why $\mathrm{rk}\, \phi = \mathrm{rk\, \psi}$ holds in this case?)

Answer 2

Since the maps $\phi,\psi$ are linear, so is $\psi\circ \phi$.

To show (i) implies (ii) . . .

Suppose $\psi\circ \phi$ is an isomorphism.

We want to show $W=\text{im}(\phi)+\text{ker}(\psi)$ and $\text{im}(\phi)\cap \text{ker}(\psi)=\{0\}$.

First we show $W=\text{im}(\phi)+\text{ker}(\psi)$.

The inclusion $\text{im}(\phi)+\text{ker}(\psi)\subseteq W$ is immediate.

For the reverse inclusion, let $w\in W$.

Since $\psi\circ \phi$ is an isomorphism, $\psi\circ \phi$ is surjective, hence $\psi(\phi(v))=\psi(w)$ for some $v\in V$.

Let $a=\phi(v)$ and let $b=w-a$.

Clearly $a\in\text{im}(\phi)$.

Also, we have $\psi(a)=\psi(\phi(v))=\psi(w)$, hence $$\psi(b)=\psi(w-a)=\psi(w)-\psi(a)=0$$ so $b\in\text{ker}(\psi)$.

Thus we have $w=a+b\in \text{im}(\phi)+\text{ker}(\psi)$.

It follows that $W\subseteq \text{im}(\phi)+\text{ker}(\psi)$.

Thus we have both inclusions, so $W=\text{im}(\phi)+\text{ker}(\psi)$.

Next we show $\text{im}(\phi)\cap \text{ker}(\psi)=\{0\}$ . . .

Suppose $w\in \text{im}(\phi)\cap \text{ker}(\psi)$.

We want to show $w=0$.

Since $w\in \text{im}(\phi)$, we have $w=\phi(v)$, for some $v\in V$.

Since $w\in \text{ker}(\psi)$, we have $\psi(w)=0$, hence $$(\psi\circ\phi)(v)=\psi(\phi(v))=\psi(w)=0$$ so $v\in \text{ker}(\psi\circ\phi)$.

Then since $\psi\circ\phi$ is an isomorphism, we get $v=0$, so $w=\phi(v)-\phi(0)=0$, as required.

It follows that $\text{im}(\phi)\cap \text{ker}(\psi)=\{0\}$.

To show (ii) implies (i) . . .

Assume $W=\text{im}(\phi)+\text{ker}(\psi)$ and $\text{im}(\phi)\cap \text{ker}(\psi)=\{0\}$.

To show $\psi\circ \phi$ is an isomorphism, if suffices to show that $\psi\circ \phi$ is bijective.

First we show $\psi\circ \phi$ is injective . . . Let $v\in \text{ker}(\psi\circ \phi)$.

We want to show $v=0$.

Let $w=\phi(v)$,

Clearly $w\in\text{im}(\phi)$.

Also from $v\in \text{ker}(\psi\circ \phi)$, we get $$\psi(w)=\psi(\phi(v))=(\psi\circ \phi)(v)=0$$ so $w\in\text{ker}(\psi)$.

Thus $w\in \text{im}(\phi)\cap \text{ker}(\psi)$, hence $w=0$.

Then $\phi(v)=w=0$, hence, since $\phi$ is injective, we get $v=0$.

It follows that $\psi\circ \phi$ is injective.

Next we show $\psi\circ \phi$ is surjective . . .

Let $x\in X$.

We want to show $x=(\psi\circ\phi)(v)$ for some $v\in V$.

Since $\psi$ is surjective, we can write $x=\psi(w)$ for some $w\in W$.

By assumption, we can write $w=a+b$, where $a\in\text{im}(\phi)$ and $b\in\text{ker}(\psi)$.

Since $a\in\text{im}(\phi)$, we can write $a=\phi(v)$ for some $v\in V$, so $$ (\psi\circ\phi)(v) = \psi(\phi(v)) = \psi(a) = \psi(a)+\psi(b) = \psi(a+b) = \psi(w) = x $$ hence $\psi\circ\phi$ is surjective.

Thus $\psi\circ\phi$ is both injective and surjective, so $\psi\circ\phi$ is an isomorphism.

This completes the proof.

Notes:

• The arguments I gave did not make any assumptions about dimensionality, and did not make any use of the Rank-Nullity Theorem.$\\[6pt]$
• I supplied a lot more detail than was strictly necessary, but I hope that doing so will flesh out the thought process. The argument can always be reduced as desired to match a preferred style.