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Let $V,W$ be vector spaces. Let's define a category whose objects are linear map $f:V\to W$ and morphisms from $f$ to $g$ are pair of linear maps $(\alpha,\beta)$ where $\alpha:V\to V,\beta :W\to W$ such that $g \circ\alpha = \beta \circ f$.

  1. Does this category have some special name?
  2. If $f_i$ are indexed family of objects, is there product of them in this category?
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As for this category having cartesian products: it's a very strange question, but the answer is 'no', unless both $V$ and $W$ are trivial ($0$-dimensional).

The basic problem can be illustrated by asking: does this category have a terminal object? Suppose $f: V \to W$ is terminal. This means that for any $h: V \to W$, there is exactly one pair of morphisms $g_1: V \to V$, $g_2: W \to W$ such that $g_2 \circ h = f \circ g_1$. Of course, by taking $g_1$ and $g_2$ both to be zero maps, there is always at least one such pair. But suppose we consider when $h: V \to W$ is the zero map. In that case every pair $(0, g_2)$ works. Thus we get uniqueness of the pair only if $W$ is trivial. But then $f = 0$, and now every pair $(g_1, 0)$ works, and we get uniqueness only if also $V$ is trivial.

The basic principle is that this category of linear maps admits products insofar as the full subcategory consisting of the single object $V$ (and similarly, the one for $W$) has products. For example, the category of linear maps has binary products if and only if we have isomorphisms $\phi: V \to V \times V$ and $\psi: W \to W \times W$, which is the case iff $V$ and $W$ are infinite-dimensional (let's agree to ignore cases where $V$ or $W$ is trivial). I.e., the composites

$$p_i = (V \stackrel{\phi}{\to} V \times V \stackrel{\pi_i}{\to} V), i = 1, 2$$

give product projections $p_1, p_2: V \to V$ in the full subcategory determined by $V$. If $q_1, q_2: W \to W$ are similar product projections for $W$, then the pairs $(p_1, q_1)$, $(p_2, q_2)$ furnish product projections for the category of linear maps. This is not hard to check. Of course, by cardinality considerations, we cannot have arbitrary products for any full subcategory determined by $V$ (again, except for the trivial case), since there will be no isomorphism $V \to V^S$ for a sufficiently high power $S$, and this carries over to the category of linear maps as well.

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  • $\begingroup$ But are you sure that the product does not exists? I tried it with similar arguments, but I failed to show it. $\endgroup$ – Gobi Oct 5 '13 at 18:56
  • $\begingroup$ Yeah, pretty sure. I can edit in an argument later about that (a little tied up with something else at the moment). $\endgroup$ – user43208 Oct 5 '13 at 19:23
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This is the morphism category that is associated to any category. If you notice, the morphisms in the morphism category are natural transformations.

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    $\begingroup$ It's often called the arrow category. ncatlab.org/nlab/show/arrow+category $\endgroup$ – Martin Brandenburg Oct 5 '13 at 11:53
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    $\begingroup$ About the product, objects of the category are linear maps from $V$ to $W$, where $V,W$ are fixed. So Isn't it different from product of vector spaces? The product of $f_i:V \to W$ should be also a linear map $f:V\to W$. $\endgroup$ – Gobi Oct 5 '13 at 12:06
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    $\begingroup$ @Gobi I hadn't realized that $V$ and $W$ were fixed. I thought you were just saying that your category was morphisms between any two vector spaces. I will remove that from my answer. $\endgroup$ – Joe Johnson 126 Oct 5 '13 at 13:08

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