# Is a right inverse of a surjective linear map linear?

On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear?

How about $T:V\rightarrow W$ linear surjection in general?

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A "surjective linear map must be an isomorphism". For example, the map from $\mathbb R^{207}$ to $\{0\}$, right? – Andrés E. Caicedo Aug 24 '12 at 3:09
@AndresCaicedo: thanks for your comment, i think i should put W=V. – Timothy Aug 24 '12 at 3:32

No. Let $V = \text{span}(e_1, e_2, ...)$ and let $T : V \to V$ be given by $T e_1 = 0, T e_i = e_{i-1}$. A right inverse $S$ for $T$ necessarily sends $v = \sum c_i e_i$ to $\sum c_i e_{i+1} + c_v e_1$ but $c_v$ may be an arbitrary function of $v$.
Let $T$ be any map of $V$ to $W$ that is onto but not one-to-one. A right inverse for $T$ is any $R: W \to V$ such that $T(R(x)) = x$ for every $x \in W$. In particular, for any $x$ such that there are $y_1 \ne y_2$ with $T(y_1) = T(y_2) = x$, you are free to make $R(x) = y_1$ or $R(x) = y_2$. If $x = u + v$ (and neither $u$ nor $v$ is $0$, so neither is $x$), at least one of these choices will not be $R(u) + R(v)$. So there will always be a nonlinear right inverse.