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consider the commutative diagram of group homomorphisms: $$\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow{g}&&\downarrow{k}\\ C&\stackrel{h}{\rightarrow}&D \end{matrix} $$

suppose $B$, $C$ and $D$ are trivial groups $\{e\}$ does this imply that necessarely $A$ is also trivial?

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    No, $A$ can be anything you want and this still commutes.2011-06-07

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Note that the following answer is just an explanation of Qiaochu's answer in more concrete terms:

Let $x\in A$. We know that $k(f(x))=h(g(x))=e$ ($e$ is the identity of $D$) since the group $D$ is trivial (and, in particular, there can be only one choice for each of $k(f(x))$ and $h(g(x))$). Therefore, the diagram commutes if $D$ is trivial. In particular, we can take $A$ to be any non-trivial group and the diagram will commute irrespective of the choices of $B$ and $C$.

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    @Theo Buehler: thankyou very much that's what i was looking for!2011-06-07
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This diagram always commutes if $D$ is trivial, regardless of the other contents of the diagram. This is because the trivial group is the terminal object in the category of groups, and maps to the terminal object are unique.

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    You might want to add that $A$ is trivial if and only if the diagram is a pull-back.2011-06-07
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    Dear Qiaochu, do you think that this answer is appropriate for such a simple question? I do not mean to suggest that this is not a good answer; in fact, this point of view can be useful in a number of general situations. However, at least in my opinion, this is overkill for someone who is thinking about this kind of question.2011-06-07
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    @Amitesh: I don't think this is overkill at all. In this context (exact sequences, commutative diagrams) it is essential to realize the (elementary) fact that there is precisely one map into and from $\{e\}$. This is used all the time. That category theorists have reserved a name (terminal/initial object) for such things, is just convenient, or at least worth knowing.2011-06-07
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    @wildildildlife Every concept in mathematics is worth knowing. The point is not which concepts are worth knowing; rather which concepts are the most important to know in the context of this question. Qiaochu said that the reason the diagram commutes (when $D$ is trivial) is because the trivial group is the terminal object in the category of groups. Well, yes, it is but this is like describing a bear as a member of the family Ursidae. The OP would need to look up the notions of a category, of a morphism in a category, which are basic to some, but not to a person who cannot solve the question.2011-06-08
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    @Amitesh: in the context of this question it is very important to know that $\{e\}$ is a terminal object. I agree with you that you don't need to call it that way, you don't need the category theoretic language. Perhaps you would have been happier if Qiaochu had added the phrase "in this case, it means that for every group $G$ there's precisely one arrow into $G\to \{e\}$."2011-06-08
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    @wildildildlife This is exactly what I had in mind. It would definitely have been better if Qiaochu added what it means for $\{e\}$ to be the terminal object in the category of groups. I agree that it is very important to know that $\{e\}$ is the terminal object (in the category of groups); my objection was that the *name* "terminal object" should be *supplemented* with some explanation. It should not itself *be* an explanation.2011-06-08
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    @Amitesh: why? The explanation on the Wikipedia page is perfectly clear.2011-06-09
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    Dear Qiaochu, perhaps we have differing views on this. I just felt (and this is, of course, a matter of opinion) that it would be difficult for someone who was thinking about this kind of problem to gain very much from the Wikipedia article. The article is clear, I suppose, as are most (mathematics) Wikipedia articles but, at least to me, this is only true if one has at least heard of the notion of a category. I guess my objection rests on the assumption that if someone does know a thing about categories and also does not know that a "morphism" in a category is "what it should be", then ...2011-06-10
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    ... the introduction to the Wikipedia article would look like a different language. But again, this is based only on my (limited) experience with people asking questions such as this one. Since you obviously have more experience on this forum compared to me, my comment probably should not be taken too seriously.2011-06-10