TITLE: The colimit of all finite-dimensional vector spaces
QUESTION [11 upvotes]: Let $\mathsf{iFinVect}_K$ be the category of finite-dimensional vector spaces with injective linear maps and $X : \mathsf{iFinVect}_K \to \mathsf{Vect}_K$ be the inclusion functor. Then $\mathrm{colim}(X)$ exists. This is because $\mathsf{Vect}_K$ is cocomplete and $\mathsf{iFinVect}_K$ is essentially small (although it is not small).
This colimit seems to be a bit strange to me, though. We merge all finite-dimensional vector spaces into a single large vector space. The embeddings are natural with respect to all injective linear maps between finite-dimensional vector spaces. Can we make this vector space more explicit? Is it, perhaps, even a well-known object? Can we find a basis?
Every element of $\mathrm{colim}(X)$ should have the form $\iota_V(v)$ for some finite-dimensional vector space $V$ and some vector $v \in V$, where $\iota_V : V \to \mathrm{colim}(X)$ is the colimit inclusion. This is because for every $V,W \in \mathsf{iFinVect}_K$ there is some $U \in \mathsf{iFinVect}_K$ with morphisms $V \xrightarrow{f} U \xleftarrow{g} W$, namely the coproduct. This implies $\iota_V(v)+\iota_W(w) = \iota_U(f(v)+g(w))$. Of course we have $\lambda \cdot \iota_V(v)=\iota_V(\lambda \cdot v)$ for $\lambda \in K$. This shows how to calculate with elements of $\mathrm{colim}(X)$.
Notice, however, that $\mathsf{iFinVect}_K$ is not filtered (because the only parallel morphisms which may be coequalized by some morphism are already equal). For this reason I think that a priori it is not so easy to decide when two elements of a colimit, say $\iota_V(v)$ and $\iota_W(w)$, are equal. It suffices to find a criterion when some element $\iota_V(v)$ is zero. This can happen when $v \neq 0$! For example, we have $0=\iota_V(v)+\iota_V(-v) = \iota_{V \oplus V}((v,-v))$.
So probably we should first answer: Do we have $\mathrm{colim}(X) \neq 0$?

REPLY [10 votes]: The colimit is the zero space.  
Let us first prove it when the field $K$ is not the field $\mathbb{F}_2$ with two elements.  Pick a scalar $\lambda \in K \setminus \{0,1\}$. For any finite-dimensional vector space $V$ and any element $v$ of $V$, the scaling map $\lambda:V\to V$ yields the equality $i_V(v)=i_V(\lambda v) = \lambda i_V(v)$.  Thus $i_V(v) = 0$.  Since this is true for any element of any vector space, the colimit has to be zero.
Now, when $K=\mathbb{F}_2$, let $V$ be any $\mathbb{F}_2$-vector space of dimension $2$ or higher, and let $v$ be any non-zero element of $V$.  There exists a unique linear map $f:\mathbb{F}_2\to V$ sending $1$ to $v$.  Thus $i_{\mathbb{F}_2}(1) = i_V(f(1)) = i_V(v)$.  This shows that all non-zero elements of $V$ have the same image under $i_V$.  This is only possible if $i_V$ is the zero map.  This proves that the colimit is again the zero space.  (Note that this argument works for any field).