Epimorphisms – Serlo

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

Motivation and derivation

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 $V,W$ be two $K$-vector spaces over the same field $K$ and $\{v_{1},\ldots ,v_{n}\}\subseteq V$ a generator. Now, what properties must a linear map $f:V\to W$ satisfy, in order for $\{f(v_{1}),\ldots ,f(v_{n})\}$ being a generator of vector space $W$? For this, we would need to be able to represent any $w\in W$ as a linear combination of $f(v_{i})$. That is, we need to find $\lambda _{1},\ldots ,\lambda _{n}\in K$ such that

$w=\sum _{i=1}^{n}\lambda _{i}f(v_{i}).$

Since the map $f$ is linear, this is equivalent to

$w=f\left(\sum _{i=1}^{n}\lambda _{i}v_{i}\right).$

So $w$ must be in the image of $f$. This is said to hold for every $w\in W$. Thus $f(V)=W$ is a necessary condition for $f$ to preserve generators.

Is this also a sufficient condition? Let $f(V)=W$. We investigate whether every $w\in W$ can be represented as a linear combination of $f(v_{i})$. Because $f(V)=W$ we have for any $w\in W$ a vector $v\in V$ with $f(v)=w$. Since $v_{1},\ldots ,v_{n}$ is a generator of $V$, there are some linear combination factors $\lambda _{1},\ldots ,\lambda _{n}$ with

$v=\sum _{i=1}^{n}\lambda _{i}v_{i}.$

So we can write $w$ as:

$w=f\left(\sum _{i=1}^{n}\lambda _{i}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}f(v_{i})$

And hence $w$ is within the generated space of the $f(v_{i})$.

Thus, the linear map $f$ preserves generators if and only if $f(V)=W$. Moreover, $f$ satisfies $f(V)=W$ exactly if $f$ is surjective. Thus, a linear map must be surjective to have the generating property. We call surjective linear maps epimorphisms.

Definition

Definition (Epimorphism)

An epimorphism is a surjective linear map $f\colon V\to W$ between two $K$-vector spaces $V$ and $W$. That is: to every $w\in W$ there is a $v\in V$ such that $w=f(v)$.

Equivalent characterization of epimorphisms

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:

Theorem

Let $f\colon V\to W$ be a linear map and let $\lbrace v_{1},\dots ,v_{n}\rbrace \subseteq V$ be a generator of $V$.

The linear map $f$ is an epimorphism exactly if $\lbrace f(v_{1}),\dots ,f(v_{n})\rbrace$ is a generator of $W$.

Proof

Proof step: „$f$ is an epimorphism“ $\implies$ „$\lbrace f(v_{1}),\dots ,f(v_{n})\rbrace$ is a generator“

Let $w\in W$ beliebig. Then according to precondition there is a vector $v\in V$ with $f(v)=w$. Since $\lbrace v_{1},\ldots ,v_{n}\rbrace$ generates the vector space $V$, there are linear combination factors $\lambda _{1},\ldots ,\lambda _{n}$ with $v=\sum _{i=1}^{n}\lambda _{i}v_{i}$. Hence, we have:

${\begin{aligned}w=f(v)=f\left(\sum _{i=1}^{n}\lambda _{i}v_{i}\right)=\sum _{i=1}^{n}f\left(\lambda _{i}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}f(v_{i}).\end{aligned}}$

So $w\in W$ can be represented as a linear combination of $f(v_{1}),\ldots ,f(v_{n})$. Since $w$ was arbitrary, $f(v_{1}),\ldots ,f(v_{n})\rbrace$ is a generator of $W$.

Proof step: „$\lbrace f(v_{1}),\dots ,f(v_{n})\rbrace$ is a generator“ $\implies$ „$f$ is an epimorphism“

Let $w\in W$ be arbitrary. We have to show that there is a vector $v\in V$ with $f(v)=w$. Since $W$ is generated by $\{\,f(v_{1}),\ldots ,f(v_{n})\,\}$, there exist scalars $\lambda _{1},\ldots ,\lambda _{n}\in K$ (for linear combination) with $w=\sum _{i=1}^{n}\lambda _{i}f(v_{i})$. We now set $v:=\sum _{i=1}^{n}\lambda _{i}v_{i}$:

${\begin{aligned}f(v)=f\left(\sum _{i=1}^{n}\lambda _{i}v_{i}\right)=\sum _{i=1}^{n}f\left(\lambda _{i}v_{i}\right)=\sum _{i=1}^{n}\lambda _{i}f(v_{i})=w.\end{aligned}}$

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

Now we generalize to vector spaces of arbitrary dimension:

Theorem

Let $f\colon V\to W$ be a linear map and let $M\subseteq V$ be a generator of $V$.

The linear map $f$ is an epimorphism if and only if $f(M)$ is a generator of $W$.

Proof

We can almost copy the proof from above: Since $M\subseteq V$ is a generator of $V$, this means that every vector $v\in V$ has a representation as a linear combination $v=\sum _{i=1}^{n}\lambda _{i}v_{i}$, where $\lambda _{1},\ldots ,\lambda _{n}$ are scalars and $v_{1},\ldots ,v_{n}$ are from $M$.

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 $1$ to $n$. Here, the number of summands depends on the vectors $v$ and $w$, 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 $f\colon V\to W$ be a homomorphism. Then the following statements are equivalent

1. $f$ is an epimorphism.
2. For all vector spaces $U$ and all $a,b\colon W\to U$ with $a\circ f=b\circ f$ we have that $a=b$. One also says that the epimorphism can be "right shortened".

Proof (Epimorphisms can be right shortened)

Proof step: 1.$\implies$2., by direct proof

Let $f\colon V\to W$ be an epimorphism, i.e., $f$ is surjective. Let $U$ be a vector space, and $a,b\colon W\to U$, such that $a\circ f=b\circ f$. We want to show that $a=b$ holds. Since $a$ and $b$ are maps with same domain of definition $W$ and same range $U$, we need to show that $a(w)=b(w)$ holds for all $w\in W$.

So consider any $w\in W$. Since $f$ is surjective, there exists a $v\in V$ with $f(v)=w$. Now, $a(w)=a(f(v))=(a\circ f)(v)=(b\circ f)(v)=b(f(v))=b(w)$. Since we have chosen $w$ arbitrary, we obtain $a=b$.

Proof step: 2.$\implies$1., proof by contradiction

Let $f\colon V\to W$ be a homomorphism. Suppose $f$ is not an epimorphism, i.e., not surjective. Then there is a $w\in W$ with $w\notin \operatorname {im} f$. In particular, $w\neq 0$ since $0=f(0)\in \operatorname {im} f$. We extend $\lbrace w\rbrace$ to a basis $B$ of $W$.

now , let us define two homomorphisms $a,b\colon W\to W$. First we set $a:=\operatorname {id} _{W}$. Further, we define $b$ using the principle of linear continuation on the basis $B$: $b\colon v\mapsto 0,b\mapsto b$ for all $b\in B\setminus \{v\}$.

Next we show $a\circ f=b\circ f$: Consider some $v\in V$. Then $f(v)\in \operatorname {span} (B\setminus \{v\})$, since $w\notin \operatorname {in} f$. As $a|_{\operatorname {span} (B\setminus \{v\})}=\operatorname {Id} _{\operatorname {span} (B\setminus \{v\})}=b|_{\operatorname {span} (B\setminus \{v\})}$, we indeed have $a(f(v))=b(f(v))$.

But $a\neq b$, since $a(w)=w\neq 0=b(w)$.

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

Examples

Example

We consider the vector spaces $\mathbb {R} ^{n}$ and $\mathbb {R} ^{m}$ with $n\geq m$, as well as the linear map

${\begin{aligned}f\colon \mathbb {R} ^{n}&\to \mathbb {R} ^{m},\\({\color {blue}x_{1},x_{2},\ldots ,x_{m}},x_{m+1},\ldots ,x_{n})&\mapsto ({\color {blue}x_{1},x_{2},\ldots ,x_{m}}).\end{aligned}}$

For this map, we simply truncate the last $n-m$ components. This makes it clear why we must require $n\geq m$ (if $n=m$, the map is simply the identity). This map is an epimorphism: Let $({\color {blue}x_{1},x_{2},\ldots ,x_{m}})\in \mathbb {R} ^{m}$. Then, we have $f({\color {blue}x_{1},x_{2},\ldots ,x_{m}},\underbrace {0,\dots ,0} _{n-m{\text{ zeros}}})=({\color {blue}x_{1},x_{2},\ldots ,x_{m}})$.

Example

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

${\begin{aligned}f:V\oplus W&\to V,\\(v,w)&\mapsto v\end{aligned}}$

Here $V\oplus W$ denotes the Outer direct sum (missing).

We first show for this that the map $f$ is linear. Consider some $(v,w),(v',w')\in V\oplus W$, as well as $\lambda \in K$. Then $f((v,w)+(v',w'))=f((v+v',w+w'))=v+v'=f((v,w))+f((v',w'))$ and $f(\lambda \cdot (v,w))=f((\lambda v,\lambda w))=\lambda v=\lambda f((v,w))$. This establishes linearity.

Let now $v\in V$ be arbitrary. Then $f((v,w))=v$ for every $w\in W$. That is, $(v,w)$ is a preimage of $v$ under $f$. Thus $f$ is an epimorphism. If $W$ is not the null space, then there are even multiple (perhaps infinitely many) preimages.

Exercises

Exercise

We consider $\mathbb {R} ^{3}$. By $e_{1}:=(1,0,0),e_{2}:=(0,1,0),e_{3}:=(0,0,1)$, we denote the standard basis. Let $f:\mathbb {R} ^{3}\to \mathbb {R} ^{2}$ be the unique linear map given by the principle of linear continuation and

${\begin{aligned}f(e_{1})={\begin{pmatrix}1\\1\end{pmatrix}},\,f(e_{2})={\begin{pmatrix}1\\-1\end{pmatrix}},\,f(e_{3})={\begin{pmatrix}1\\-1\end{pmatrix}}\end{aligned}}$

Show that $f$ is an epimorphism.

Solution

By construction, $f$ is a linear map. We want to show that $\{f(e_{1}),f(e_{2}),f(e_{3})\}=\{(1,1),(1,-1)\}$ is a generator of $\mathbb {R} ^{2}$. Then, we have with the above theorem that $f$ is an epimorphism. So, we must represent every $(x,y)\in \mathbb {R} ^{2}$ as a linear combination of vectors $(1,1),(1,-1)$. Accordingly, we search linear combination scalars $\lambda ,\mu \in \mathbb {R}$ such that

${\begin{aligned}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}1\\1\end{pmatrix}}+\mu {\begin{pmatrix}1\\-1\end{pmatrix}}.\end{aligned}}$

From this we get the linear system of equations

${\begin{aligned}{\begin{aligned}x&=\lambda +\mu \\y&=\lambda -\mu ,\end{aligned}}\end{aligned}}$

which is solved by $\lambda ={\frac {x+y}{2}}$ and $\mu ={\frac {x-y}{2}}$. So we have ${\begin{pmatrix}x\\y\end{pmatrix}}={\frac {x+y}{2}}{\begin{pmatrix}1\\1\end{pmatrix}}+{\frac {x-y}{2}}{\begin{pmatrix}1\\-1\end{pmatrix}}$ for all $x,y\in \mathbb {R}$. Thus $\{(1,1),(1,-1)\}$ is a generator. This proves that $f$ is an epimorphism.

Exercise

Consider the function space $\operatorname {Fun} ([0,1],\mathbb {R} )$ of all function s mapping $\mathbb {R}$ to $\mathbb {R}$, as well as the map

${\begin{aligned}\phi \colon \operatorname {Fun} ([0,1],\mathbb {R} )&\to \mathbb {R} ,\\f&\mapsto f(0).\end{aligned}}$

Show that $\phi$ is an epimorphism.

Solution

The operations on the function space are defined element-wise in each case. That is: for $f,g\in \operatorname {Fun} ([0,1],\mathbb {R} )$, $\lambda \in \mathbb {R}$ and $x\in [0,1]$ we have that $(f+g)(x)=f(x)+g(x)$ and $(\lambda f)(x)=\lambda f(x)$. In particular, this is true for $x=0$, which implies

${\begin{aligned}\phi (f+g)=(f+g)(0)=f(0)+g(0)=\phi (f)+\phi (g)\end{aligned}}$

and

${\begin{aligned}\phi (\lambda f)=(\lambda f)(0)=\lambda f(0)=\lambda \phi (f)\end{aligned}}$

Thus we have shown the linearity.

To prove surjectivity, let $c\in \mathbb {R}$ be arbitrary. We need to show that there is a $f:[0,1]\to \mathbb {R}$ with $f(0)=c$. Such a map exists, since e.g. the constant function

${\begin{aligned}f:[0,1]\to \mathbb {R} ,\,x\mapsto c\end{aligned}}$

has the desired properties. Every $c\in \mathbb {R}$ thus has a preimage, so $\phi$ is an epimorphism.