I anticipated there would be many answers to this already, but couldn't find any. I am aware of this question: , and I think I can deduce the answer from it, but I am very unsure, so I prefer a direct answer.  Let me best give an example. Recently, I noticed a sharp decline in the (last) price of the stock S. I placed a limit order for 10 stocks with a limit EUR 51,74. I got them on the next day at the opening for exactly EUR 51,50.  So, my question is: How does the price at which I buy really come about. Is it the cheapest selling price (ask) compliant with me order (meaning sellers always loose), is it the mean between my limit and the ask? Is it something entirely different?  Conversely, if I sell the stock with a limit order at EUR 55,00, does it mean I will only get 55,00, although there are people bidding EUR 56,00?  If it depends on the location, answers for XETRA are most appreciated.  EDIT  I think, that the suggested duplicate covers me, but I would like to be certain I understand correctly.  1)I got the stock for EUR 51,50, merely because the sellers ask predated my bid.  2)If the sellers ask of, say. EUR 51,50 had come after my bid, he would have gotten the full 51,74.  EDIT   I try to provide any information desired. Because I saw the comments, I add the following. I apologize, should it be irrelevant. I am (obviously) a newbie.  I ordered on one day shortly after 17:30 using XETRA. I chose the validity option  "today" and the option "continuous trade". I was not aware of the stop of trading after 17:30.   The system (German) said that my order was received and would be presented ("vorgetragen") on the next day. Then, on the next day in the morning I got the stock for EUR51,50.
Let's assume a very simplified situation. There is only one seller asking for 10$ and there is only one buyer bidding 20$. At what price would the transaction go?   Some answers say it won't, because the ask price has to match the bid price. () Other answers say, that the time ordering is important. () Which answer is correct? What is the final price in the presented situation?
In a preadditive category, products are coproducts $\renewcommand{\phi}{\varphi}$ $\require{AMScd}$ My question is different from .  I want to see how in a preadditive category $\mathcal{C}$ a product is a coproduct, following . I'll write the proof up to the point where I get stuck.  Let $A=A_1,B=A_2$ be objects and suppose their product $P$ does exist, with projections $\pi_i:P\rightarrow A_i$. By considering a wedge  $$\begin{CD}A @&lt;{id_A}&lt;&lt;  A @&gt;{0_{AB}}&gt;&gt;  B\end{CD}$$  and the obvious analogous one with $A$ and $B$ exchanged, where $0_{A,B}$ is the zero morphism, we see that the product property implies the existence of two unique morphisms $k_i\in\mathcal{C}[A_i,P]$ s.t.  \begin{align*} \pi_j\circ k_i=\begin{cases} \mathrm{id}_i &amp;\text{if }i=j\\ 0_{A_i, A_j} &amp;\text{if }i\neq j \end{cases}. \end{align*} Using these equations and the definition of zero morphism, it's easy to see that the morphism $m\equiv k_1\circ \pi_1+ k_2\circ \pi_2$ satisfies $\pi_i\circ m=\pi_i$.   This is where I get stuck: In the linked proof, they now state that this implies $m=\mathrm{id}_P.$ But I don't see why this should be true.   Specifically, if $\phi\in\mathcal{C}[S,P]$ for some arbitrary object $S$, it is not apparent why $m\circ\phi=\phi$. 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