Surjectivity Asks Whether Every Output Can Be Reached
A linear map has a domain, where its inputs come from, and a codomain, where its outputs are required to lie. The map may not actually produce every vector in that codomain. Surjectivity is the condition that it does. To decide whether a map is surjective, focus on the outputs it can produce: its image. In examples, this amounts to asking whether the equation \(T(\mathbf{v})=\mathbf{w}\) can be solved for each target \(\mathbf{w}\) in the codomain.
Throughout, the vector spaces and linear maps in the examples are over \(\mathbb{R}\). For \(n\in\mathbb{N}\), \(P_n\) denotes the vector space of real polynomials of degree at most \(n\), including the zero polynomial. An element of \(P_n\) has the form \(a_0+a_1x+\cdots+a_nx^n\), with some coefficients possibly zero.
The image is always contained in the codomain, since every output of \(T:V\to W\) belongs to \(W\). The key question is whether this containment is equality. The definition of the image records exactly which outputs are attained, so it gives the central characterization.
Proof. Suppose \(T\) is surjective. For any \(\mathbf{w}\in W\), surjectivity gives a vector \(\mathbf{v}\in V\) with \(T(\mathbf{v})=\mathbf{w}\). By the definition of image, this means \(\mathbf{w}\in\operatorname{im}T\), so \(W\subseteq\operatorname{im}T\). Every value of \(T\) belongs to \(W\), so \(\operatorname{im}T\subseteq W\). The two inclusions give \(\operatorname{im}T=W\). Conversely, suppose \(\operatorname{im}T=W\). For every \(\mathbf{w}\in W\), the equality implies \(\mathbf{w}\in\operatorname{im}T\). By the definition of image, some \(\mathbf{v}\in V\) satisfies \(T(\mathbf{v})=\mathbf{w}\). This is precisely surjectivity. \(\square\)
Testing Polynomial Outputs
For a polynomial map, a proposed output is attainable if it equals the value of the map for some polynomial in the domain. To show that a map is onto, one must be able to produce a preimage for an arbitrary polynomial in the codomain. To show that it is not onto, it is enough to find a single codomain polynomial with no preimage.
Worked Example: Differentiation from \(P_2\) to \(P_1\)
Define \(D:P_2\to P_1\) by \(D(p)=p'\), the derivative of \(p\). Determine whether \(D\) is surjective, and find a preimage of \(q(x)=3-4x\).
Take an arbitrary target \(q(x)=a+bx\in P_1\). We seek \(p(x)=c_0+c_1x+c_2x^2\in P_2\) satisfying \(p'(x)=a+bx\). Differentiating gives
Matching coefficients requires \(c_1=a\) and \(2c_2=b\), so \(c_2=b/2\). The coefficient \(c_0\) can be any real number. In particular, choose \(c_0=0\). Then \(p(x)=ax+\tfrac{b}{2}x^2\) is in \(P_2\), and
Every polynomial in the codomain has a preimage, so \(D\) is surjective. For the specified target \(q(x)=3-4x\), take \(a=3\) and \(b=-4\). The preimage given by the formula is \(p(x)=3x-2x^2\).
Check: \(D(3x-2x^2)=3-4x\), as required.
This example depends on the declared codomain. Differentiating a polynomial in \(P_2\) always produces a polynomial in \(P_1\), but that does not mean the same rule reaches every polynomial in \(P_2\). The codomain is part of the map, not just a label that can be ignored.
Worked Example: Differentiation from \(P_2\) to \(P_2\)
Now define \(S:P_2\to P_2\) by \(S(p)=p'\). Is \(S\) surjective? In particular, can the output \(q(x)=x^2\) be attained?
For every input \(p(x)=c_0+c_1x+c_2x^2\in P_2\),
The coefficient of \(x^2\) in every output is zero. The target \(x^2\in P_2\) has coefficient \(1\) on \(x^2\), so it cannot equal \(S(p)\) for any \(p\in P_2\). Therefore \(x^2\notin\operatorname{im}S\), and \(\operatorname{im}S\ne P_2\). The map \(S\) is not surjective.
Check: A formal attempt to solve \(p'=x^2\) would require \(p(x)=\tfrac13x^3+c_0\). For any choice of \(c_0\), this is not in \(P_2\), so it is not an allowed input to \(S\).
Testing Coordinate-Space Outputs
For a map between coordinate spaces, write a general input, apply the defining formula, and compare the result with a target. If the target has arbitrary coordinates, a formula that constructs an input from those coordinates can prove surjectivity. If the output coordinates are constrained in a way some codomain vectors violate, the map is not onto.
Worked Example: Every Vector in \(\mathbb{R}^2\) Is Reached
Define \(T:\mathbb{R}^3\to\mathbb{R}^2\) by \(T(x,y,z)=(x+z,y-z)\). Test whether an arbitrary target \((a,b)\in\mathbb{R}^2\) is attained, then find a preimage of \((4,-1)\).
To solve \(T(x,y,z)=(a,b)\), we need \(x+z=a\) and \(y-z=b\). Choose \(z=0\); the equations then give \(x=a\) and \(y=b\). Thus the input \((a,b,0)\in\mathbb{R}^3\) works for every target \((a,b)\in\mathbb{R}^2\):
Every codomain vector is attained, so \(T\) is surjective. For the particular target \((4,-1)\), this construction gives the preimage \((4,-1,0)\).
Check: \(T(4,-1,0)=(4+0,-1-0)=(4,-1)\).
Worked Example: A Coordinate Relation Prevents Surjectivity
Define \(U:\mathbb{R}^2\to\mathbb{R}^3\) by \(U(s,t)=(s+t,2s+2t,0)\). Describe its image and decide whether the target \((1,0,0)\) is attained.
For an input \((s,t)\), set \(r=s+t\). Then \(U(s,t)=(r,2r,0)\), so every output has second coordinate twice its first coordinate and third coordinate zero. Conversely, for any \(r\in\mathbb{R}\), the input \((r,0)\) gives \(U(r,0)=(r,2r,0)\). Hence the image is exactly
The target \((1,0,0)\) would require \(s+t=1\) and \(2s+2t=0\). The first equation gives \(2s+2t=2\), contradicting the second. Thus \((1,0,0)\) is not attained. Since it belongs to the codomain \(\mathbb{R}^3\), but not to the image, \(U\) is not surjective.
Check: The image description also excludes \((1,0,0)\): if its first coordinate were \(r=1\), the second coordinate would have to be \(2\), not \(0\).
A Finite Test Using a Basis
When the codomain is finite-dimensional, it is not necessary to check every vector in it separately. A basis is a list that is linearly independent and spans the space; spanning means every vector in the space is a linear combination of the list. For a linear map, attaining every vector in a basis is enough to ensure that every linear combination of those vectors is attained.
Proof. If \(T\) is surjective, every vector in \(W\) has a preimage. In particular, for each \(i\), there is some \(\mathbf{v}_i\in V\) with \(T(\mathbf{v}_i)=\mathbf{w}_i\), so \(\mathbf{w}_i\in\operatorname{im}T\). For the converse, suppose each basis vector has a preimage: choose \(\mathbf{v}_i\in V\) with \(T(\mathbf{v}_i)=\mathbf{w}_i\). Let \(\mathbf{w}\in W\) be arbitrary. Because \(\mathcal{B}\) is a basis, there are scalars \(c_1,\dots,c_m\) such that \(\mathbf{w}=c_1\mathbf{w}_1+\cdots+c_m\mathbf{w}_m\). Linearity gives \(T(c_1\mathbf{v}_1+\cdots+c_m\mathbf{v}_m)=c_1T(\mathbf{v}_1)+\cdots+c_mT(\mathbf{v}_m)=c_1\mathbf{w}_1+\cdots+c_m\mathbf{w}_m=\mathbf{w}\). Thus the arbitrary target \(\mathbf{w}\) has a preimage, and \(T\) is surjective. \(\square\)
For the map \(T:\mathbb{R}^3\to\mathbb{R}^2\) above, the standard basis of \(\mathbb{R}^2\) is \((\mathbf{e}_1,\mathbf{e}_2)\), where \(\mathbf{e}_1=(1,0)\) and \(\mathbf{e}_2=(0,1)\). The formula gives \(T(1,0,0)=(1,0)=\mathbf{e}_1\) and \(T(0,1,0)=(0,1)=\mathbf{e}_2\). Both basis vectors are attained, so the theorem confirms that \(T\) is onto. The proof explains why this finite check is valid: linear combinations of preimages are preimages of the corresponding linear combinations of outputs.
Common Pitfall: One Attained Output Is Not Enough
A common error is to find one target that has a preimage and conclude that the map is surjective. Consider \(F:\mathbb{R}^2\to\mathbb{R}^2\) given by \(F(s,t)=(s,0)\). The target \((1,0)\) is attained, since \(F(1,0)=(1,0)\). But that single calculation does not show that every target is attained. In fact, a mistaken claim that \(F\) is onto fails at \((0,1)\): an output of \(F\) always has second coordinate \(0\), so the equation \(F(s,t)=(0,1)\) has no solution.
The correction is to match the claim with the required test. To prove surjectivity, construct a preimage for an arbitrary target, or—for a finite-dimensional codomain—prove that every vector in a basis has a preimage. To disprove surjectivity, one unattained codomain vector is enough. Always check that the proposed target lies in the stated codomain, and keep that codomain fixed while determining the image.
Check Your Understanding
For each question, distinguish the target space from the outputs the map actually produces.
- For \(D:P_2\to P_1\) defined by \(D(p)=p'\), find a preimage of \(2+6x\) and verify it by differentiating.
- For \(S:P_2\to P_2\) defined by \(S(p)=p'\), explain why \(1+x^2\) has no preimage.
- For \(G:\mathbb{R}^2\to\mathbb{R}^2\) given by \(G(s,t)=(s+t,s-t)\), solve for a preimage of an arbitrary target \((a,b)\). Is \(G\) surjective?
- For \(H:\mathbb{R}^2\to\mathbb{R}^3\) given by \(H(s,t)=(s,t,s+t)\), give a codomain vector that is not attained and explain why.
- State and prove the basis test for surjectivity when the codomain has a finite basis.
