Extension and Lift: A Factoring Duality
Extension and Lift: A Factoring Duality
Extension (shared domain)
Suppose $ W_1 $ is finite-dimensional, $ S \in \mathcal{L}(V, W_1) $ and $ T \in \mathcal{L}(V, W_2) $. Then \(\operatorname{null} S \subseteq \operatorname{null} T \iff \exists\, E \in \mathcal{L}(W_1, W_2) \ \text{s.t.}\ T = ES.\)
($\Leftarrow$) If $T = ES$ and $Sv = 0$, then $Tv = E(Sv) = E0 = 0$, so $\operatorname{null} S \subseteq \operatorname{null} T$.
($\Rightarrow$) We build $E : W_1 \to W_2$ with $E(Sv) = Tv$ for all $v$.
For $w \in \operatorname{range} S$, write $w = Sv$ and set $Ew := Tv$. This is well-defined: if $Sv_1 = Sv_2$, then $v_1 - v_2 \in \operatorname{null} S \subseteq \operatorname{null} T$, so $Tv_1 = Tv_2$. It is linear on $\operatorname{range} S$ since, writing $w_i = Sv_i$, \(E(aw_1 + bw_2) = E\bigl(S(av_1+bv_2)\bigr) = T(av_1+bv_2) = aEw_1 + bEw_2.\)
Because $W_1$ is finite-dimensional, $\operatorname{range} S$ has a complement: \(W_1 = \operatorname{range} S \oplus U.\) Define $E$ to be $0$ on $U$ and extend linearly to all of $W_1$. Then $Ew \in W_2$ for every $w$, and $ES = T$. $\blacksquare$
Lift (shared codomain)
Suppose $ V_1 $ is finite-dimensional, $ S \in \mathcal{L}(V_1, W) $ and $ T \in \mathcal{L}(V_2, W) $. Then \(\operatorname{range} S \subseteq \operatorname{range} T \iff \exists\, E \in \mathcal{L}(V_1, V_2) \ \text{s.t.}\ S = TE.\)
($\Leftarrow$) If $S = TE$, then $Sv = T(Ev) \in \operatorname{range} T$ for all $v$, so $\operatorname{range} S \subseteq \operatorname{range} T$.
($\Rightarrow$) Because $V_1$ is finite-dimensional, pick a basis $v_1, \dots, v_n$ of $V_1$. For each $j$, the vector $Sv_j$ lies in $\operatorname{range} S \subseteq \operatorname{range} T$, so there exists $u_j \in V_2$ with \(T u_j = S v_j.\) (The preimage $u_j$ lives in $V_2$ because $T$ starts at $V_2$ — which is exactly why $E$ ends up in $\mathcal{L}(V_1, V_2)$.) Define $E \in \mathcal{L}(V_1, V_2)$ to be the unique linear map with $E v_j = u_j$ for every $j$. Then for each basis vector, \((TE)(v_j) = T(E v_j) = T(u_j) = S v_j.\) So $TE$ and $S$ agree on a basis of $V_1$, hence $TE = S$. $\blacksquare$
Extension and Lift are transposes.
The two laws are one statement seen through the dual map $M \mapsto M’$. Transpose the Extension data — $S : V \to W_1$ and $T : V \to W_2$, sharing the domain $V$ — to get $S’ : W_1’ \to V’$ and $T’ : W_2’ \to V’$, now sharing the codomain $V’$, which is a Lift configuration. The conclusion transposes with it, $T = ES \mapsto T’ = S’E’$, turning “$T$ factors through $S$ on the left” into “$T’$ factors through $S’$ on the right.”
And the hypotheses correspond: since $\operatorname{range} M’ = (\operatorname{null} M)^0$, and since $U \mapsto U^0$ is an inclusion-reversing bijection on subspaces (its inverse is $U^0 \mapsto U^{00} = U$, the double-annihilator identity), \(\operatorname{null} S \subseteq \operatorname{null} T \iff (\operatorname{null} T)^0 \subseteq (\operatorname{null} S)^0 \iff \operatorname{range} T' \subseteq \operatorname{range} S'.\) So the Extension problem for $(S, T)$ is exactly the Lift problem for its transposes $(T’, S’)$.
Removing the finite-dimensional hypotheses
Both laws hold in every dimension under the Axiom of Choice; the finite-dimensional hypotheses are conveniences, not necessities.
- Extension used $W_1$ finite-dimensional only to give $\operatorname{range} S$ a complement in $W_1$. Under AC, every subspace of every vector space has a complement, so the hypothesis can be dropped.
- Lift used $V_1$ finite-dimensional only to supply a basis of $V_1$ and to choose a preimage $u_j$ for each basis vector. Under AC, every vector space has a (Hamel) basis and the preimages can be chosen simultaneously, so the hypothesis can be dropped.
The reverse ($\Leftarrow$) directions use neither a complement nor a basis, so they are choice-free in all dimensions. Structurally, the two facts being invoked are that in $\mathbf{Vect}$ every object is both injective (extension along monos) and projective (lifting along epis).
Left and Right Inverse
Setting one given map to an identity collapses each law into a one-sided inverse:
Extension. Put $T = I_V$. The hypothesis $\operatorname{null} S \subseteq \operatorname{null} I_V = {0}$ says $S$ is injective, and $I_V = ES$ makes $E$ a left inverse of $S$. A left inverse is called a retraction of $S$, and an injective map that admits one is a split monomorphism.
Lift. Put $S = I_W$. The hypothesis $W = \operatorname{range} I_W \subseteq \operatorname{range} T$ says $T$ is surjective, and $I_W = TE$ makes $E$ a right inverse of $T$. A right inverse is called a section of $T$, and a surjective map that admits one is a split epimorphism.
So injective $\iff$ has a retraction and surjective $\iff$ has a section.
A retraction is automatically surjective and a section automatically injective, so the two constructions say that in $\mathbf{Vect}$ every injective map splits (has a retraction) and every surjective map splits (has a section) — the linear-algebra form of “every mono and every epi splits.”
These two equivalences are transposes of each other, inherited from the Extension↔Lift duality above.
Since setting a map to the identity commutes with transposing, the retraction↔section pairing is that duality evaluated at $I$: a retraction $ST = I_V$ dualizes to $T’S’ = I_{V’}$, a section of $T’$. Bringing in the second annihilator identity $\operatorname{null} M’ = (\operatorname{range} M)^0$ alongside the first completes the dictionary between the two maps’ properties:
\[T \text{ injective} \iff T' \text{ surjective}, \qquad T \text{ surjective} \iff T' \text{ injective}.\]So the transpose swaps injective with surjective in the very stroke that swaps retraction with section: “$T$ is injective, hence has a retraction” read on the dual side is exactly “$T’$ is surjective, hence has a section.” The left/right asymmetry is one phenomenon, seen through $M \mapsto M’$.
| relationship | factorization | $E$ acts on |
|---|---|---|
| $\operatorname{range} S \subseteq \operatorname{range} T$ | $S = TE$, $E \in \mathcal{L}(V)$ | domain (one-directional) |
| $\operatorname{null} S \subseteq \operatorname{null} T$ | $T = ES$, $E \in \mathcal{L}(W)$ | codomain (one-directional) |
| $\operatorname{range} S = \operatorname{range} T$ | $S = TE$, $E \in \mathcal{L}(V)$ invertible | domain (isomorphism) |
| $\operatorname{null} S = \operatorname{null} T$ | $S = ET$, $E \in \mathcal{L}(W)$ invertible | codomain (isomorphism) |