0
$\begingroup$

Let $C$ be a category. Let $X,Y\in\mathcal{C}$ be two objects and $Z\in\mathcal{C}$ be the final object. I want to show $X\times_Z Y\cong X\times Y$.

So there is only 1 morphism in each $Mor(X,Z)$ and $Mor(Y,Z)$. Call the corresponding morphisms $f:X\to Z$ and $g:Y\to Z$. Thus one can define $X\times_ZY$ for given $f,g$. Furthermore $X\times_ZY\to X\times Y$ is possible by $X\times_Z Y$ factoring through $X\times Y$ as there are two projection mappings $\pi_1:X\times_ZY\to X$ and $\pi_2:X\times_ZY\to Y$ or by universal property of products.

Since there $X\times Y\in Obj(C)$, the morphism $X\times Y\to Z$ is defined by $f\circ\pi_1$ and $g\circ\pi_2$. However, $Z$ is final. There is only one element in $Mor(X\times Y, Z)$. Thus $f\circ\pi_1=g\circ\pi_2$. So $X\times Y$ satisfies fiber product universal property with maps $f,g$. By fiber product being final, $X\times Y\cong X\times_Z Y$ as all final objects are isomorphic.

  • 0
    You'd use the universal product of $X\times_Z Y$. The whole exercise can be viewed as just showing that $X \times Y$ has the universal property of the fiber product in this case. Once that's established the isomorphism is the general isomorphism between object satisfying the same universal property. – 2017-01-30
  • 0
    @DerekElkins. That is what I thought. However, there is no reason to believe given $(x,y)\in X\times Y$, $f(x)=g(y)$. So that is the reason I doubt this part. Was I thinking this part wrong? – 2017-01-30
  • 0
    Can you describe the universal property of the *final* object? – 2017-01-30
  • 0
    @DerekElkins .Any $Z$ is final if $|Mor(X,Z)|=1$ for any $X\in Obj(C)$. This determines maps $f:X\to Z$ and $g:Y\to Z$ uniquely. If I take $f^{-1}(z)\times g^{-1}(z)$, I may get $\phi\times $something non-empty. I do not know what I am missing here. I feel the conclusion should be $f(x)=g(y)$ for $(x,y)\in X\times Y$ to make this work. – 2017-01-30
  • 0
    This is not correct to write $(x,y) \in X \times Y$ for a general category $\mathcal C$. But it can lead your intuition, so assume you're in $\mathsf{Set}$ for now. In which case can you have $f(x) \neq g(y)$ (remember the codomain of $f,g$) ? – 2017-01-30
  • 0
    $f(x)$ and $g(y)$ mean $f \circ \pi_1$ and $g \circ \pi_2$. These are two arrows $X\times Y \to Z$, but the universal property of the final object says that's there's only one arrow, so they must be equal. – 2017-01-30
  • 0
    @DerekElkins ,@Pece Thanks a lot. I think I got the right idea and I have corrected the question statement with the answer. – 2017-01-30

0 Answers 0