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 $\mathrm{colim}(X) \neq 0$?

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    $\begingroup$ Perhaps I'm missing something, but doesn't multiplication by a non-zero and non-unity scalar $\lambda:V\to V$ yield that $i_V(v) = i_V(\lambda(v)) = \lambda i_V(v)$? This would imply that $i_V(v)$ is zero, provided that the field is not $\mathbb{F}_2$. $\endgroup$ Apr 22 '15 at 9:26
  • $\begingroup$ You are right! You can make this into an answer. The case of $\mathbb{F}_2$ has to be considered separately. $\endgroup$ Apr 22 '15 at 9:41
  • $\begingroup$ 6 upvotes for a question about a colimit which results in ZERO. $\endgroup$ Apr 22 '15 at 9:43
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    $\begingroup$ The construction is probably the most intricate way I've seen to describe the zero space, though :) $\endgroup$ Apr 22 '15 at 9:46
  • $\begingroup$ I think there is a typo : shouldn't it be $i_V(v) + i_W(w) = i_U(f(v)+g(w))$ ? $\endgroup$
    – Pece
    Apr 22 '15 at 10:55

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).

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    $\begingroup$ I guess that I was too fixated on the idea that the colimit is interesting. Meanwhile I've also found that the colimit of $\mathsf{iFinSet} \to \mathsf{Set}$ is boring. $\endgroup$ Apr 22 '15 at 13:45
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    $\begingroup$ @MartinBrandenburg That of $\mathsf{iFinGrp} \to \mathsf{Grp}$ looks promising, though. $\endgroup$ Apr 22 '15 at 13:54
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    $\begingroup$ @Pierre-GuyPlamondon That's also boring. Much the same argument you gave for vector spaces shows that $i_G(g)=i_G(h)$ if $g,h\in G$ have the same order. So for $G$ simple, $i_G$ is trivial, and any finite group embeds in a finite simple group. $\endgroup$ Apr 22 '15 at 15:32
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    $\begingroup$ @JeremyRickard: It surprises me that your argument is not so "simple". Probably there are more direct proofs? $\endgroup$ Apr 22 '15 at 16:33
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    $\begingroup$ @MartinBrandenburg A more elementary, though maybe longer, proof, using no knowledge of particular groups except cyclic ones: By essentially the same argument as Pierre-Guy's, for each $n$ there is an element $g_n$ of the colimit that is the image of all elements, of all finite groups, of order $n$. Since the order of any element is the same as that of its inverse, every $g_n$ has order at most $2$. But $g_n=g_{2n}^2$, so $g_n$ is the identity. $\endgroup$ Apr 23 '15 at 8:46

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