Two Different Ways to Change a Linear Map
Let \(T:V\to W\) be a linear map. We can change the map’s domain by choosing a subspace \(U\) of \(V\) and considering only inputs in \(U\). We can also change its codomain: if all outputs of \(T\) lie in a subspace \(C\) of \(W\), the same rule can be viewed as a map from \(V\) to \(C\). These operations look similar, but their effects are different. Restricting the domain can change both the kernel and image. Changing only the codomain leaves the kernel untouched, though it can change whether the map is surjective.
Throughout, the spaces are vector spaces over a field \(\mathbb{F}\), and the maps are linear. The kernel of \(T\) is \(\ker T=\{\mathbf{v}\in V:T(\mathbf{v})=\mathbf{0}\}\), and its image is \(\operatorname{im}T=\{T(\mathbf{v}):\mathbf{v}\in V\}\). A map is injective if equal outputs imply equal inputs; it is surjective if every element of its codomain is an output.
The restriction is well-defined because every \(\mathbf{u}\in U\) is an input in \(V\), so \(T(\mathbf{u})\in W\). The corestriction is well-defined precisely because the condition \(\operatorname{im}T\subseteq C\) ensures that each output belongs to its new codomain. Both maps are linear: they use the same rule as \(T\), and the addition and scalar multiplication in a subspace are inherited from the original space.
Restricting the Domain
When the domain is reduced from \(V\) to \(U\), only those inputs in \(U\) remain available. The outputs are consequently \(T(U)=\{T(\mathbf{u}):\mathbf{u}\in U\}\), which may be smaller than \(\operatorname{im}T\). The inputs in \(U\) that map to zero are exactly the inputs shared by \(U\) and the original kernel.
Proof. An element \(\mathbf{u}\in U\) belongs to \(\ker(T|_U)\) exactly when \(T|_U(\mathbf{u})=\mathbf{0}\). Since \(T|_U(\mathbf{u})=T(\mathbf{u})\), this is equivalent to \(\mathbf{u}\in\ker T\). Together with \(\mathbf{u}\in U\), this gives \(\ker(T|_U)=U\cap\ker T\). By the definition of image, the outputs of \(T|_U\) are exactly the vectors \(T(\mathbf{u})\) for \(\mathbf{u}\in U\), so \(\operatorname{im}(T|_U)=T(U)\). A linear map is injective exactly when its kernel contains only the zero vector, so the kernel equality gives the stated injectivity condition. Finally, \(T|_U\) is surjective onto \(W\) exactly when every vector in \(W\) is \(T(\mathbf{u})\) for some \(\mathbf{u}\in U\); this is precisely \(T(U)=W\). \(\square\)
The injectivity condition says that the chosen subspace must avoid all nonzero vectors in the original kernel. It does not require the original map to have been injective: the restriction may discard the inputs responsible for its failure of injectivity. Surjectivity onto \(W\), on the other hand, requires the inputs in \(U\) to produce every vector in the original codomain. A smaller domain can make that harder, but it can still be surjective if its outputs cover \(W\).
Worked Example: Restricting a Projection to a Diagonal Line
Define \(T:\mathbb{R}^2\to\mathbb{R}^2\) by \[ T\left(\begin{bmatrix}x\\y\end{bmatrix}\right)=\begin{bmatrix}x\\0\end{bmatrix}, \] and let \(U=\operatorname{span}\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}\). Determine whether \(T|_U:U\to\mathbb{R}^2\) is injective and surjective.
Every vector in \(U\) has the form \(\begin{bmatrix}a\\a\end{bmatrix}\), with \(a\in\mathbb{R}\). Applying the map gives \[ T\left(\begin{bmatrix}a\\a\end{bmatrix}\right)=\begin{bmatrix}a\\0\end{bmatrix}. \] This output is zero exactly when \(a=0\). Thus \(\ker(T|_U)=\{\mathbf{0}\}\), and the restriction is injective. Its image is \(\left\{\begin{bmatrix}a\\0\end{bmatrix}:a\in\mathbb{R}\right\}\), the horizontal axis, which is not all of \(\mathbb{R}^2\). For example, \(\begin{bmatrix}0\\1\end{bmatrix}\) is not an output, so the restriction is not surjective onto \(\mathbb{R}^2\).
Check: If two restricted inputs have equal outputs, then \[ \begin{bmatrix}a\\0\end{bmatrix} = \begin{bmatrix}b\\0\end{bmatrix} \quad\Longrightarrow\quad a=b. \] So their inputs \(\begin{bmatrix}a\\a\end{bmatrix}\) and \(\begin{bmatrix}b\\b\end{bmatrix}\) are equal, confirming injectivity. The second coordinate of every output is zero, confirming the failure of surjectivity.
The same original map can behave differently on another subspace. Its kernel in \(\mathbb{R}^2\) is the vertical axis, so choosing that axis as the restricted domain retains, rather than removes, the nonzero kernel vectors.
Worked Example: Restricting to the Original Kernel
Use the same map \(T\), but now let \(U=\operatorname{span}\left\{\begin{bmatrix}0\\1\end{bmatrix}\right\}\). Find the kernel and image of \(T|_U\), and classify its injectivity and surjectivity.
An arbitrary input in \(U\) is \(\begin{bmatrix}0\\a\end{bmatrix}\). Then \[ T\left(\begin{bmatrix}0\\a\end{bmatrix}\right) = \begin{bmatrix}0\\0\end{bmatrix} \quad\text{for every }a\in\mathbb{R}. \] Hence every vector of \(U\) is in the restricted kernel, so \(\ker(T|_U)=U\), and its image is \(\{\mathbf{0}\}\). Since \(\begin{bmatrix}0\\1\end{bmatrix}\ne\mathbf{0}\) is in the kernel, the restriction is not injective. Since its image does not contain \(\begin{bmatrix}1\\0\end{bmatrix}\), it is not surjective onto \(\mathbb{R}^2\).
Check: The theorem gives \(U\cap\ker T=U\), because \(U\) is exactly the vertical axis, and gives \(T(U)=\{\mathbf{0}\}\). These are the kernel and image computed directly.
Changing the Codomain
Suppose \(C\) is a subspace of \(W\) that contains every output of \(T\). Passing from \(T:V\to W\) to \(T_C:V\to C\) does not alter the inputs or the output rule. In particular, it cannot change which inputs map to zero. It can change surjectivity because the new codomain may be smaller: to be surjective onto \(C\), the image must equal \(C\).
Proof. For each \(\mathbf{v}\in V\), the value \(T_C(\mathbf{v})\) is the same vector as \(T(\mathbf{v})\). Therefore \(T_C(\mathbf{v})=\mathbf{0}\) if and only if \(T(\mathbf{v})=\mathbf{0}\), which proves \(\ker T_C=\ker T\). The set of outputs is unchanged as well, so \(\operatorname{im}T_C=\operatorname{im}T\), now viewed inside \(C\). Since injectivity is equivalent to having only the zero vector in the kernel, the equality of kernels proves the injectivity claim. By definition, \(T_C\) is surjective onto \(C\) exactly when its image is \(C\), so this is equivalent to \(\operatorname{im}T=C\). Taking \(C=\operatorname{im}T\) satisfies the containment hypothesis and makes the image equal to the codomain, proving the final claim. \(\square\)
More generally, if both changes are made at once, the map \(T|_U:U\to C\) is defined whenever \(T(U)\subseteq C\). Its kernel is \(U\cap\ker T\), and its image is \(T(U)\), regarded as a subspace of \(C\). It is injective exactly when \(U\cap\ker T=\{\mathbf{0}\}\), and surjective onto \(C\) exactly when \(T(U)=C\). These conditions concern different features: avoiding the kernel controls injectivity, while producing the whole stated codomain controls surjectivity.
Worked Example: Corestricting a Projection to Its Image
Define \(S:\mathbb{R}^3\to\mathbb{R}^3\) by \[ S\left(\begin{bmatrix}x\\y\\z\end{bmatrix}\right) = \begin{bmatrix}x\\y\\0\end{bmatrix}. \] Find its image and kernel, then classify the original map and its corestriction \(S_{\operatorname{im}S}:\mathbb{R}^3\to\operatorname{im}S\).
Every output has third coordinate zero, and every vector \(\begin{bmatrix}a\\b\\0\end{bmatrix}\) occurs by taking input \(\begin{bmatrix}a\\b\\0\end{bmatrix}\). Therefore \[ \operatorname{im}S = \left\{\begin{bmatrix}a\\b\\0\end{bmatrix}:a,b\in\mathbb{R}\right\}. \] The output is zero exactly when \(x=0\) and \(y=0\), with \(z\) unrestricted. Thus \[ \ker S = \left\{\begin{bmatrix}0\\0\\z\end{bmatrix}:z\in\mathbb{R}\right\}. \] The original map is not surjective onto \(\mathbb{R}^3\), since \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) is not in its image. It is not injective, since the nonzero vector \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) is in its kernel.
For the corestriction, the codomain is exactly the plane \(\operatorname{im}S\). Every \(\begin{bmatrix}a\\b\\0\end{bmatrix}\) in this codomain is the output of the input \(\begin{bmatrix}a\\b\\0\end{bmatrix}\), so the corestriction is surjective. Its kernel remains \(\ker S\), so it is still not injective.
Check: \[ S\left(\begin{bmatrix}a\\b\\0\end{bmatrix}\right) = \begin{bmatrix}a\\b\\0\end{bmatrix}, \qquad S\left(\begin{bmatrix}0\\0\\1\end{bmatrix}\right) = \begin{bmatrix}0\\0\\0\end{bmatrix}. \] The first equality verifies surjectivity onto the image plane; the second verifies that the nonzero kernel vector remains after corestriction.
Common Pitfall: A Smaller Codomain Does Not Remove the Kernel
A common incorrect conclusion is that corestricting a map to its image makes it injective as well as surjective. The example above directly disproves this: after corestricting \(S\) to its image plane, the nonzero input \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) still maps to \(\begin{bmatrix}0\\0\\0\end{bmatrix}\). The mistake is to confuse the outputs available in the codomain with the inputs that map to zero. Corestriction changes the first description, not the output rule or the kernel.
There is a related pitfall in domain restriction. The original map \(T:\mathbb{R}^2\to\mathbb{R}^2\), \(T(x,y)=(x,0)\), is not injective because \(T(0,1)=(0,0)\). But on the diagonal line \(U=\operatorname{span}\{(1,1)\}\), the only vector also in \(\ker T\) is zero, so the restriction is injective. Saying that a map’s injectivity never changes under domain restriction ignores the possibility of discarding nonzero kernel vectors. The exact test is \(U\cap\ker T=\{\mathbf{0}\}\).
These distinctions are useful whenever a map is studied only on a chosen family of inputs, or when its actual output space is more informative than a larger declared codomain. The domain determines which inputs are tested and which outputs can be reached. The codomain determines the target against which surjectivity is judged. Keeping these roles separate makes it possible to state precisely what has changed—and what has not.
Check Your Understanding
For each question, identify the domain and codomain being used before deciding injectivity or surjectivity.
- If \(U\le V\), express \(\ker(T|_U)\) using \(U\) and \(\ker T\).
- For \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(x,0)\), is the restriction to \(\operatorname{span}\{(1,1)\}\) injective? Is it surjective onto \(\mathbb{R}^2\)?
- State the condition on a subspace \(C\le W\) that makes the corestriction \(T_C:V\to C\) well-defined.
- Explain why corestricting a map to \(\operatorname{im}T\) always makes it surjective, but does not necessarily make it injective.
- For a map \(T|_U:U\to C\) with \(T(U)\subseteq C\), state separate conditions for injectivity and surjectivity.
