5

I came across a result on the page Theorems implied by Yoneda's lemma? which said that Yoneda's Lemma implies the isomorphism of the title; namely, if we have $\text{Hom}(X,Z)\cong \text{Hom}(Y,Z)$, then $X\cong Y$. However my knowledge of category theory is limited at best. I was wondering if anyone could explain how this works, or under what conditions (if any) it fails to work? In particular I want to use it for modules whose underlying abelian groups are simply copies of $\mathbb{Z}$

Community
  • 1
Sam Williams
  • 1,401
  • 3
    You want that to be true for every choice of $Z$, not only one $Z$. It is easy to come up with examples, say $X=\Bbb Z_4$, $Y=\Bbb Z_2$, $Z=\Bbb Z$. Both homs are zero. – Pedro Oct 16 '14 at 21:02
  • 1
    It also has to be a natural bijection, or at least, natural enough. – Zhen Lin Oct 16 '14 at 22:02
  • Related: https://math.stackexchange.com/questions/704891 – Watson Sep 23 '17 at 13:19

1 Answers1

4

Let me show how a family of isomorphisms $\mathcal{C}(X,Z)\cong\mathcal{C}(Y,Z)$ natural in $Z$ gives us an isomorphism $Y\cong X$. I will assume that you are familiar with functors, in particular the $\operatorname{Hom}$-funcotrs in question and natural transformations, because otherwise discussing implications of Yoneda's Lemma won't be very fruitful.

For $\mathcal{C}$ locally small, we have the Yoneda-Functor $\mathcal{C}^{\operatorname{op}}\xrightarrow{\ \ \mathcal{Y}\ \ }[\mathcal{C},\mathbf{Set}]$ sending an arrow $Y\xrightarrow{\ \ f\ \ }X$ on the natural transformation $$\left(\mathcal{C}(X,Z)\xrightarrow{k\longmapsto kf}\mathcal{C}(Y,Z)\right)_{Z\in\mathcal{C}}\,.$$ If we can show that this functor is fully faithful, it will follow that every isomorphism $\mathcal{C}(X,-)\cong\mathcal{C}(Y,-)$ comes from a unique isomorphism $Y\cong X$.

Now the Yoneda Lemma states that given any functor $\mathcal{C}\xrightarrow{\ \ F\ \ }\mathbf{Set}$ and any object $X\in\mathcal{C}$ the map $$FX\longrightarrow[\mathcal{C},\mathbf{Set}](\mathcal{C}(X,-),F)\,,\qquad x\longmapsto(\mathcal{C}(X,Z)\xrightarrow{k\longmapsto (Fk)x}FZ)_{Z\in\mathcal{C}}$$ is a bijection (natural in both $F$ and $X$ for an appropriate functor).

For $F=\mathcal{C}(Y,-)$ this means that $$\mathcal{C}(Y,X)\longrightarrow[\mathcal{C},\mathbf{Set}](\mathcal{C}(X,-),\mathcal{C}(Y,-))\,,\qquad f\longmapsto \left(\mathcal{C}(X,Z)\xrightarrow{k\longmapsto \mathcal{C}(Y,k)(f)}\mathcal{C}(Y,Z)\right)_{Z\in\mathcal{C}}$$ is a bijection. But since $\mathcal{C}(Y,k)(f)=kf$, this is exactly the map $\mathcal{C}(Y,X)\longrightarrow[\mathcal{C},\mathbf{Set}](\mathcal{Y}X,\mathcal{Y}Y)$, $f\longmapsto\mathcal{Y}f$, so the Yoneda functor is indeed full and faithful.