The colimit of all finite-dimensional vector spaces

Solution 1:

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).