First Isomorphism Theorem of quotient space

Question

Let $T:V\to W$ be a linear map, define $S:V/\ker(T) \to \text{Im}(T)$ as $S(v+\ker(T))=T(v)$. Show that $S$ defines an isomorphism.

My professor gave us the proof where it involves showing $S$ is well-defined, linear, injective and surjective one by one. Am I right in thinking both injectivity and subjectivity are not required as a linear map that is injective must be also surjective by Rank-Nullity. Will this in turn make this proof slightly redundant?

(Also a rookie question, do we always need to check well-definedness when defining a new map?)

Answer 1

That is not true, a linear map can be injective without being surjective. For example, consider $f:\Bbb{R}^2 \rightarrow \Bbb{R}^3$ given by $$f(\mathbf{x})=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\mathbf{x}.$$ Then $f$ is linear, injective (because rank is $2$) but definitely not surjective (e.g. $\begin{bmatrix}0\\0\\1\end{bmatrix}$ is not in the range).

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For your second question, to check well-definedness. It is always important to make sure that the function is well-defined and it becomes even more important when the domain you are dealing with has multiple representations of the same object (e.g. set of equivalence classes).

Answer 2

Consider the map $\mathbb R^3\to\mathbb R^4$ given by $(1,0,0)\mapsto (1,0,0,0)$, $(0,1,0)\mapsto (0,1,0,0)$, and $(0,0,1)\mapsto(0,0,1,0)$. This is injective but not surjective.

Answer 3

This isomorphism theorem is stronger than rank-nullity, because in the isomorphism theorem, we do not assume that $V$ is finite-dimensional, whereas we do assume this in the rank-nullity theorem. So you cannot use rank-nullity to prove the isomorphism theorem, but you can use the isomorphism theorem to prove rank-nullity. (i.e the isomorphism theorem can be viewed as a generalization of rank-nullity to infinite-dimensions)

By the way, your statement

"a linear map that is injective must also be surjective by rank-nullity"

is strictly speaking false. The correct statement is:

Let $V$ and $W$ be vector spaces over a field $F$, with equal finite dimension, and let $T: V \to W$ be linear. Then, $T$ is injective if and only if $T$ is surjective.

(and the proof makes use of rank-nullity).

I suggest you re-read all the theorems and pay close attention to all the dimension-related hypotheses, and where they used in the proofs.

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As a general rule of thumb, whenever you're defining things on quotient spaces, you need to check well-definition, because the objects you're dealing with are equivalence classes. So, if you use a particular representative, you need to check that using another representative still gives the same result.

So, for example, you need to check well-definition when you:

• define addition and scalar multiplication on quotient spaces
• define maps between quotient spaces