# A question on the composition of relations. [closed]

Suppose that $f$ is a relation from $A$ to $B$ and that $g$ is a relation from $B$ to $C$ such that $\operatorname{Dom}(f) = A$ and $\operatorname{Dom}(g) = B$. Prove that if $g \circ f:A \to C$ (that is, $g \circ f$ is a function from $A$ to $C$) and $\operatorname{Rng}(f) = B$, then $g:B \to C$ (that is, $g$ is a function from $B$ to $C$).

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## closed as too localized by cardinal, Benjamin Lim, The Chaz 2.0, t.b., Asaf KaragilaApr 20 '12 at 23:33

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I don't understand the downvotes. Yes, there's not exactly an effort made to show own work on this, but there are bazillions of other questions that don't do that either. It may not be the most advanced question ever either, but we're not at mathoverflow here. –  Desiato Apr 20 '12 at 1:01
@Desiato: You may want to look at the meta thread and the list of the OP's questions. My sense are that the downvotes are trying to send a strong message to the OP that his behavior lies outside the expected norms of this community. So far the message appears not to have arrived to its intended destination...yet. –  cardinal Apr 20 '12 at 1:30
I read the meta thread. I just don't see anything in this question that lies "outside the expected norms", as far as I can observe. –  Desiato Apr 20 '12 at 2:23
@Desiato: Copying and pasting homework with no effort to solve it themselves is generally considered unacceptable behavior, especially when done repeatedly. You might also be interested in the meta threads concerning best practices for answering homework questions; they get special treatment on this site. –  cardinal Apr 20 '12 at 3:16
@cardinal: I think this is kinda becoming meta here (and I don't know how to move to chat), but I'm not sure how else to reply to you. If I understand Miguel's response on meta correctly, he and his buddy are trying to study for an exam as opposed to doing homework. It's a bit difficult to read since he obviously is esl. But exactly due to that point, I would hope for them to be accepted with a little bit of tolerance here, compared to the average question. –  Desiato Apr 20 '12 at 3:22

To show that $g: B\to C$ is a function, we need to show that $dom(g)=B$ and for all $b \in B$ there is exactly one $c \in C$ with $(b,c) \in g$.
$dom(g)=B$: Let $b \in B$. Since $ran(f)=B$, there is $a \in A$ with $f(a)=b$ and since $g \circ f$ is a function, there is $c \in C$ with $(a,c)\in g\circ f$, and thus $(b,c)\in g$.
Now assume that there is $(b,c_1), (b,c_2) \in g$. Since $f$ is a function with $ran(f)=B$, there is $a\in A$ with $(a,b) \in f$. Since $g \circ f$ is a function, there is a unique image of $a$, and thus $c_1 = c_2$.