Epimorphisms – Serlo

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Linear maps preserve linear combinations. We now learn about special linear maps that preserve generators. These are called epimorphisms.

Motivation and derivation[Bearbeiten]

In the article on monomorphisms we considered linear maps which map linearly independent vectors to linearly independent vectors. There we found out that these maps are exactly injective linear maps. Injective linear maps therefore "preserve" linear independence.

Using the linear independence, we could express the intuitive dimension notion in mathematical (linear algebra) terms. There, we also encountered generators. Now: Are there also linear maps that preserve generators?

So let be two -vector spaces over the same field and a generator. Now, what properties must a linear map satisfy, in order for being a generator of vector space ? For this, we would need to be able to represent any as a linear combination of . That is, we need to find such that

Since the map is linear, this is equivalent to

So must be in the image of . This is said to hold for every . Thus is a necessary condition for to preserve generators.

Is this also a sufficient condition? Let . We investigate whether every can be represented as a linear combination of . Because we have for any a vector with . Since is a generator of , there are some linear combination factors with

So we can write as:

And hence is within the generated space of the .

Thus, the linear map preserves generators if and only if . Moreover, satisfies exactly if is surjective. Thus, a linear map must be surjective to have the generating property. We call surjective linear maps epimorphisms.


Definition (Epimorphism)

An epimorphism is a surjective linear map between two -vector spaces and . That is: to every there is a such that .

Equivalent characterization of epimorphisms[Bearbeiten]

We have already considered in the motivation that surjective linear maps are exactly the maps that preserve generators.
Because the case of finite generators is more important than the general statement, we consider this case first. Then we investigate what we need to change for the general case:


Let be a linear map and let be a generator of .

The linear map is an epimorphism exactly if is a generator of .


Proof step: is an epimorphism“ is a generator“

Let beliebig. Then according to precondition there is a vector with . Since generates the vector space , there are linear combination factors with . Hence, we have:

So can be represented as a linear combination of . Since was arbitrary, is a generator of .

Proof step: is a generator“ is an epimorphism“

Let be arbitrary. We have to show that there is a vector with . Since is generated by , there exist scalars (for linear combination) with . We now set :

This proves that is surjective, that is, an epimorphism.

Now we generalize to vector spaces of arbitrary dimension:


Let be a linear map and let be a generator of .

The linear map is an epimorphism if and only if is a generator of .


We can almost copy the proof from above: Since is a generator of , this means that every vector has a representation as a linear combination , where are scalars and are from .

The only thing that changes is that the sums no longer have a fixed number of summands. In the proof above, we could always run the sums of to . Here, the number of summands depends on the vectors and , respectively. But it is still a finite number of summands. Therefore, the rest of the proof is the same as within the finite case.

We will now be introduced to a second (category-theoretic) characterization of epimorphisms, the possibility of being "right shortened":

Theorem (Epimorphisms can be right shortened)

Let be a homomorphism. Then the following statements are equivalent

  1. is an epimorphism.
  2. For all vector spaces and all with we have that . One also says that the epimorphism can be "right shortened".

Proof (Epimorphisms can be right shortened)

Proof step: 1.2., by direct proof

Let be an epimorphism, i.e., is surjective. Let be a vector space, and , such that . We want to show that holds. Since and are maps with same domain of definition and same range , we need to show that holds for all .

So consider any . Since is surjective, there exists a with . Now, . Since we have chosen arbitrary, we obtain .

Proof step: 2.1., proof by contradiction

Let be a homomorphism. Suppose is not an epimorphism, i.e., not surjective. Then there is a with . In particular, since . We extend to a basis of .

now , let us define two homomorphisms . First we set . Further, we define using the principle of linear continuation on the basis : for all .

Next we show : Consider some . Then , since . As , we indeed have .

But , since .

This is a contradiction to the assumption, and it follows that is an epimorphism.



We consider the vector spaces and with , as well as the linear map

For this map, we simply truncate the last components. This makes it clear why we must require (if , the map is simply the identity). This map is an epimorphism: Let . Then, we have .


For a field and two -vector spaces , the following map is an epimorphism:

Here denotes the Outer direct sum (missing).

We first show for this that the map is linear. Consider some , as well as . Then and . This establishes linearity.

Let now be arbitrary. Then for every . That is, is a preimage of under . Thus is an epimorphism. If is not the null space, then there are even multiple (perhaps infinitely many) preimages.



We consider . By , we denote the standard basis. Let be the unique linear map given by the principle of linear continuation and

Show that is an epimorphism.


By construction, is a linear map. We want to show that is a generator of . Then, we have with the above theorem that is an epimorphism. So, we must represent every as a linear combination of vectors . Accordingly, we search linear combination scalars such that

From this we get the linear system of equations

which is solved by and . So we have for all . Thus is a generator. This proves that is an epimorphism.


Consider the function space of all function s mapping to , as well as the map

Show that is an epimorphism.


The operations on the function space are defined element-wise in each case. That is: for , and we have that and . In particular, this is true for , which implies


Thus we have shown the linearity.

To prove surjectivity, let be arbitrary. We need to show that there is a with . Such a map exists, since e.g. the constant function

has the desired properties. Every thus has a preimage, so is an epimorphism.