I’m not sure about my answer to the following problem:
Problem: Let $A,B$ and $C$ be sets, and let $f:A \rightarrow C$ and $g:B \rightarrow C$ be maps. Show that there exists a set $P$ and maps $h:P \rightarrow A$ and $k:P \rightarrow B$ such that $f \circ h = g \circ h$, and that for any set $X$ and maps $s:X \rightarrow A$ and $t:X \rightarrow B$ such that $f \circ s = g \circ t$, there is a unique map $u:X \rightarrow P$ such that $s = h \circ u$ and $t = k \circ u$.
Here it is my solution.
Solution: I’m dividing my solution in three parts in order to be more organised:
I started to define the set $P$ as $P = \{(x,y) \in A \times B | f(x) = g(y)\}$ and the maps $h:P \rightarrow A$ and $k:P \rightarrow B$ as $h((x,y))=x$ and $k((x,y))=y$ for all $(x,y) \in P$. Then it follows that $f \circ h, g \circ k:P \rightarrow C$. For $x \in P$, we deduce that $x = (a,b)$ with $a \in A$, $b \in B$ and $f(a)=g(y)$. So $(f \circ h)(x)=(f(h(x))=f(h((a,b)))=f(a)=g(y)=g(k((a,b)))=g(k(x))=(g \circ k)(x)$. Therefore $f \circ h = g \circ k$.
For the next step, I defined the map $u:X \rightarrow P$ as $u(x)=(s(x),t(x))$ for all $x \in X$. Now, let $x \in X$. We know that $(f \circ s)(x)=(g \circ t)(x)$, so $f(s(x))=g(t(x))$. Then $(s(x),t(x)) \in P$, which means that $u(x) \in P$. We observe that $h \circ u:X \rightarrow A$. Hence $(h \circ u)(x)=h(u(x))=h((s(x),t(x))=s(x)$. Therefore $h \circ u = s$. By the same reasoning, we conclude that $k \circ u = t$. This proves the existence of map $u$.
Now we turn our attention to the uniqueness of map $u$. Suppose that $u_1,u_2:X \rightarrow P$ are maps such that $h \circ u_1 = s = h \circ u_2$ and $k \circ u_1 = t = k \circ u_2$. Let $x \in X$, then $u_1(x) = (u_{1,1}(x),u_{1,2}(x))=((h \circ u_1)(x),(k \circ u_1)(x)) = (s(x),t(x))=((h \circ u_2)(x),(k \circ u_2)(x))=(u_{2,1}(x),u_{2,2}(x))=u_2(x)$. Therefore $u_1 = u_2$. So such map is unique.
What is concerning me on this solution is:
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Is it really necessary that $f \circ s = g \circ t$?
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Since I rarely use the above condition, I feel like something is missing in the solution (specially in the uniqueness part).
Any ideas or comments about that? Thank you for your time!
Best Answer
The thing about pullbacks in general is that it is a pair of maps $h:P\to A$ and $k:P\to B$ that are universal with the property $f\circ h=g\circ k$. In other words, it is somehow the "best pair of functions" that achieve this property. Therefore, it's necessary to compare the pair $(h,k)$ only against pairs $(s,t)$ that also satisfy this property; that is, $f\circ s=g\circ t$. The metric of being "better" is measured by the existence of a unique map $u$ through which $s$ and $t$ factor to recover $h$ and $k$ (which you've stated precisely in your question).
You mention that you "hardly use" the property $f\circ s=g\circ t$, and sure it may have been used only once, but it was used in a crucial way: the map $u:X\to P$ you defined would not exist otherwise. The map $u:X\to P$ is necessarily unique without this condition, because like you've shown, the set $P$ is a subset of $A\times B$ and so functions into $P$ are determined by their action on the components. Since $h$ and $k$ are just projections into the respective components, any two $u_1,u_2:X\to P$ that agree on components will be equal.
You can use this fact to realise the necessity of $f\circ s=g\circ t$ for the existence portion: by the uniqueness argument, you are forced to define $u:X\to P$ as $u(x) := (s(x),t(x))$ as you've done, but this is only a well-defined function $X\to P$ iff $(s(x),t(x))\in P$ for all $x$; that is, $f(s(x))=g(t(x))$ for all $x\in X$.