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Imagine a semisimple abelian category $\mathcal{C}$, for example representations of a finite group. Take two (nonsimple) objects $X, Y$ that are subobjects of a common object $Z$, and decompose them into simples, say $X \cong X_1 \oplus X_2 \oplus \cdots \oplus X_n$ and $Y \cong Y_1 \oplus Y_2 \oplus \cdots \oplus Y_m$. If some of the $X_i$ and $Y_j$ are isomorphic and are being mapped onto the same objects in $Z$, say $X_{i_k} \cong Y_{j_k}$ for $k \in \{1, \ldots K\}$, then $X$ and $Y$ have a common subobject $S$, namely $S := X_{i_1} \oplus X_{i_2} \oplus \cdots \oplus X_{i_K}$ such that $S \hookrightarrow X \hookrightarrow Z = S \hookrightarrow Y \hookrightarrow Z$. It is a maximal object with this property, all other common subobjects are subobjects of it.

This looks a lot like a universal property, but I can't figure out which one. What especially intrigues me that I'm looking not for any objects, but for subobjects. How do I encode that into the universal property?

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  • $\begingroup$ What you've written doesn't uniquely determine $S$. Think about the case that $C = \text{Vect}$. $\endgroup$
    – Qiaochu Yuan
    Commented Apr 30, 2014 at 17:38
  • $\begingroup$ (If this were a universal property, $S$ would be the terminal object in the category of triples $(S, i_X, i_Y)$ where $S$ is an object and $i_X : S \to X, i_Y : S \to Y$ are monomorphisms. You don't often see universal properties with conditions on the morphisms like that, and indeed even for an innocuous category like $\text{Vect}$ I think this category fails to have a terminal object almost all the time.) $\endgroup$
    – Qiaochu Yuan
    Commented Apr 30, 2014 at 17:50
  • $\begingroup$ @QiaochuYuan, No, I think you've misunderstood something. For the case of $\mathrm{Vect}$, you have $X = \mathbb{C}^n$ and $Y = \mathbb{C}^m$. Then $S = \mathbb{C}^{\min(n,m)}$. $\endgroup$
    – Manuel Bärenz
    Commented Apr 30, 2014 at 19:46
  • $\begingroup$ If the $X_i$ are all mutually nonisomorphic and the $Y_j$ as well, then it is, I believe, the product in the category of subobjects of the coend $\Omega_{\mathcal{C}}$, which is the direct sum of all simples. But I'm unsure about this, and I don't know how to get rid of the additional assumption. $\endgroup$
    – Manuel Bärenz
    Commented Apr 30, 2014 at 19:51
  • $\begingroup$ I mean you haven't uniquely determined $S$ as a subobject. What are the maps $S \to X, S \to Y$ in your example? $\endgroup$
    – Qiaochu Yuan
    Commented Apr 30, 2014 at 21:14

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One can define the category of subobjects of $Z$, where the objects are subobjects of $Z$ and the morphisms have to commute with the monomorphisms of the subobjects. In this category, $S$ is simply the product of $X$ and $Y$.

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