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Let $X$ be a normed space, $Y$ a Banach space, and $T:X\to Y$ a linear map. If its kernel is complete in the original norm and $S:Y\to X$ is a bounded right inverse, then the graph norm $\Vert x{\Vert }_T=\Vert x{\Vert }_X+\Vert Tx{\Vert }_Y$ is complete, even if $X$ with its original norm is incomplete or $T$ is unbounded.

For a Cauchy sequence $(x_n)$ in the graph norm, $y_n=T{x}_n$ converges to $y\in Y$. The elements $w_n=x_n-S{y}_n$ lie in the kernel and satisfy $\Vert w_n-w_m{\Vert }_X\le (1+\Vert S\Vert )\Vert x_n-x_m{\Vert }_T$. Hence $w_n\to w\in \ker T$. The element $x=w+Sy$ satisfies $Tx=y$ and

The differentiation operator sends a differentiable function to its derivative; its domain and the norms must be specified. On $C^1([-1,1])$ with values in $C([-1,1])$, using the supremum norm on both, it is an unbounded linear operator: $f_n(x)=\sin (nx)$ satisfies $\Vert f_n{\Vert }_{\infty }\le 1$ and $\Vert D{f}_n{\Vert }_{\infty }=n$. Its graph norm is $\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, which is complete. Indeed, the kernel consists of constants and is complete, while $Sg(x)={\int }_0^{x}g(t)dt$ is a bounded right inverse by the fundamental theorem of calculus; apply completeness of a graph norm from a bounded right inverse.

An inverse function reverses a bijection. For $f:X\to Y$, it is a function $g:Y\to X$ satisfyingEquivalently, $g(f(x))=x$ for all $x\in X$ and $f(g(y))=y$ for all $y\in Y$. We write $g=f^{-1}$. The superscript denotes reversal through function composition; it does not mean the pointwise multiplicative inverse $1/f(x)$. The preimage notation $f^{-1}(B)$ for a set $B$ is also distinct: a preimage exists even when no inverse function exists.

For a computed example, let $f:\mathbb{R}\to \mathbb{R}$ be $f(x)=2x+1$. Solving $y=2x+1$ for $x$ gives $g(y)=(y-1)/2$. Thus $f(1)=3$ and $g(3)=1$; direct substitution gives $g(f(x))=x$ and $f(g(y))=y$ throughout the real line. In contrast, $1/f(3)=1/7$, which does not undo $f(1)=3$.

Why must $f$ be a bijection? If $f(x_1)=f(x_2)$, applying $g$ gives $x_1=x_2$, proving injectivity. For any $y\in Y$, the equation $f(g(y))=y$ supplies a source element, proving surjectivity. Conversely, a bijection supplies exactly one predecessor for each $y$, defining a unique inverse function. A left inverse or right inverse requires only one of the two composition identities.

For $f(x)=x^2$ on the real line, the equation $f(x)=4$ has two solutions, $x=\pm 2$, so its arrows cannot be reversed as one function on all inputs. The restriction to $[0,\infty )$ removes that ambiguity. As a bijection from $[0,\infty )$ onto $[0,\infty )$, it has inverse function $g(y)=\sqrt{y}$. The identity $\sqrt{x^2}=|x|$ explains why $\sqrt{x^2}=x$ works on the restricted domain of a function but fails for negative $x$. ^[ref].

More generally, $(a,b)$ lies on the graph of a function $f$ precisely when $(b,a)$ lies on the graph of a function of its inverse function. This coordinate swap explains the geometric reflection and the horizontal line test. It also explains why inverse sensitivity is reciprocal locally: the inverse function rule divides by the original derivative at the corresponding input, when that derivative is nonzero.

A linear map $T:V\to W$ between finite-dimensional vector spaces admits a linear left inverse $S:W\to V$ if and only if its kernel of a linear map is $\{0\}$, equivalently if and only if it is injective.

Necessity follows from $Tv=0⟹v=STv=0$. For sufficiency, choose a basis $v_1,\dots ,v_n$ of $V$. Triviality of the kernel of a linear map implies that $T{v}_1,\dots ,T{v}_n$ are linearly independent. By basis extension, they extend to a basis $T{v}_1,\dots ,T{v}_n,w_{n+1},\dots ,w_m$ of $W$. Define $S(T{v}_i)=v_i$ and $S(w_j)=0$, and extend by linearity. Then $ST{v}_i=v_i$ on a basis, proving $ST=I_V$.

The freedom in $S$ lies in its values on a direct-sum complement of the image of a linear map. For example, for $T(x,y)=(x,y,0)$, every linear map $S(u,v,w)=(u+aw,v+bw)$ is a left inverse, with arbitrary scalars $a,b$. A right inverse instead requires $T$ to be a surjective linear map.

Let $(x_n)$ be a Cauchy sequence in the graph norm, and set $y_n=T{x}_n$. Since$(y_n)$ is a Cauchy sequence in the Banach space $Y$. Write $y_n\to y\in Y$.

Use the bounded right inverse to remove the part seen by $T$:Thus $w_n\in W$, and the operator norm bound givesThe assumed completeness of $W$ supplies $w\in W$ with $w_n\to w$ in $\Vert \cdot {\Vert }_X$. Put $x=w+Sy\in X$. The linearity of $T$ and the right inverse identity give $Tx=y$. ThereforeWe have proved the completeness of a graph norm from a bounded right inverse:

The kernel $W=⟨1⟩$ is complete: the map $a\mapsto a1$ is an isometry from $\mathbb{R}$ with its absolute-value norm onto $W$ with the supremum norm.

Define a linear map $S:Y\to X$ byThe fundamental theorem of calculus gives $Sg\in C^1([-1,1])$ and $TSg=g$. Also,Thus $S$ is a bounded right inverse, and $Y=C([-1,1])$ is a Banach space. Applying part (c),

A left inverse of $f$ is a function $\ell :B\to A$ satisfying $\ell \circ f={\mathrm{id}}_A$. If such $\ell$ exists and $f(a)=f(a^{\prime })$, applying $\ell$ gives $a=a^{\prime }$, so $f$ is an injection. Conversely, suppose $f$ is an injection and fix $a_0\in A$, possible because $A$ is nonempty. DefineThe first value is unique by injectivity, and $\ell (f(a))=a$ for every $a$. ThusNo choice from a family of nonempty sets is needed here: each preimage in the first case is unique, and the second case uses one fixed element.

A right inverse is a function $r:B\to A$ with $f\circ r={\mathrm{id}}_B$. Its existence makes every $b$ the image of $r(b)$, proving that $f$ is a surjection. Conversely, if $f$ is a surjection, every fiber of a function $f^{-1}(\{b\})$ is nonempty. Using the usual axiom of choice for arbitrary sets, choose $r(b)\in f^{-1}(\{b\})$ for each $b$. Then $f(r(b))=b$, givingThe converse in this second equivalence is the right-inverse characterization of the axiom of choice; it is not an unrestricted choice-free theorem about arbitrary sets.

Choose a right inverse $s:A\to B$ of the surjection $f$, so $f(s(a))=a$, and defineFor each $a$, the element $b=s(a)$ then witnesses both $f(b)=a$ and $g(b)=h(a)$. As in part (a), arbitrary simultaneous choices use the axiom of choice. If $A$ is empty, surjectivity and the existence of $f:B\to A$ force $B$ to be empty, and the unique empty function $h:A\to C$ already meets the requirement.

Suppose $g$ is constant on each fiber of a function of $f$. Since every fiber is nonempty, $h(a)$ must be its common $g$-value: any witness $b$ in that fiber gives that value. This defines exactly one $h$ and yields $g=h\circ f$, an instance of factorization through a surjection.

Conversely, suppose a fiber contains $b,b^{\prime }$ with $f(b)=f(b^{\prime })=a_0$ but $g(b)\ne g(b^{\prime })$. Starting with any admissible $h$, define $h_1,h_2$ to agree with it away from $a_0$, but put $h_1(a_0)=g(b)$ and $h_2(a_0)=g(b^{\prime })$. Both remain admissible, witnessed at $a_0$ by $b$ and $b^{\prime }$ respectively, yet they are distinct. HenceThe distinction is important: the original existence property asks only for one witness in each fiber. Only when $g$ is fiberwise constant does that property force $g=h\circ f$ for every $b$.

Take $fg=f\circ g$. The assertion can fail on an infinite set. On $X={\mathbb{N}}_0$, set $g(n)=n+1$, $f(0)=0$, and $f(n)=n-1$ for $n\ge 1$. Then $f(g(n))=n$ for every $n$, but $g(f(0))=1\ne 0$. Thus a right inverse need not be a left inverse.

On a finite set it is always true. From $fg={\mathrm{id}}_X$, $g$ is an injection: equality of two $g$-values gives equality after applying $f$. An injective self-map of a finite set is a bijection. Composing $fg={\mathrm{id}}_X$ with $g^{-1}$ shows $f=g^{-1}$, and hence $gf={\mathrm{id}}_X$. This includes the empty set.

True. Use the lifted basis $u_1,\dots ,u_r$ above and put $W=\mathrm{span}\{u_1,\dots ,u_r\}$. If $\sum c_i{u}_i=0$, applying $\alpha$ gives $\sum c_i{v}_i=0$, so all $c_i=0$. Thus the $u_i$ form a basis of $W$, and $\alpha {|}_W$ takes this basis to the basis of $V$. It is a linear isomorphism.

In fact the right inverse gives the stronger splittingFor every $u\in U$, write $u=(u-\beta \alpha u)+\beta \alpha u$; the first term is in the kernel and the second in $W$. Their intersection is zero because $\alpha {|}_W$ is injective. This is a direct sum decomposition.

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.

If $TS={\mathrm{id}}_W$, every $w\in W$ would satisfy $w=T(Sw)$, so $T$ would be a surjective linear map. But its image of a linear map has dimension at most $\dim V=2$, whereas $\dim W=3$. Therefore no such $S$ exists. This is the obstruction to a right inverse: $T$ cannot reach every target vector.

The system is consistent for every $b\in {\mathbb{R}}^m$: take $x=Yb$, since $A(Yb)=(AY)b=b$. Thus $Y$ is a right inverse giving an explicit solution for every right-hand side.

All solutions are $Yb+u$ with $u$ in the kernel of a linear map represented by $A$. The rank-nullity theorem gives kernel dimension $n-m$; the solution is unique when $n=m$, and is nonunique when $n>m$.

Suppose $g(y)=g(y^{\prime })$. Applying $f$ and using the right inverse identity givesThis is exactly injectivity, so every right inverse is injective.

The Moore--Penrose inverse is characterized by

Here $A^{†}=A^{+}$ denotes the Moore-Penrose inverse, while the star denotes the conjugate transpose. The first two identities express consistency when applying and reversing the linear map; the last two make the resulting orthogonal projection matrices Hermitian. They must hold together. For example, $A=(12)$ and $B=(10{)}^T$ satisfy $ABA=A$ and $BAB=B$, butis not Hermitian. This $B$ is a right inverse, yet $Bb=(3,0{)}^T$ is not the minimum-norm least-squares solution for $b=3$. The last two Penrose equations impose the orthogonality that selects the pseudoinverse.

The assertion that every surjective function has a right inverse is equivalent to the axiom of choice. Choice selects one member of each nonempty fiber of a function, giving a right inverse. Conversely, for a family $(X_i{)}_{i\in I}$ of nonempty sets, the projection from the disjoint union $\{(i,x):i\in I,x\in X_i\}$ onto $I$ is surjective. A right inverse selects one element from each $X_i$, giving a choice function.