When $X \times Y \cong X \times Z$ implies $Y \cong Z$ (in the category of finite topological spaces)

The monoid $\mathcal M_{\rm fin}$ is in fact cancellative.


To prove this, we start with a Lemma. Let us write $h(T,Y)$ for the cardinality of the set of continuous maps $T\to Y$ and $i(T,Y)$ for the cardinality of the set of injective continuous maps $T\to Y$.

Lemma: Let $Y$ and $Z$ be finite spaces such that $h(T,Y)=h(T,Z)$ for all finite spaces $T$. Then $i(T,Y)=i(Y,Z)$ for all finite spaces $T$.

Proof of Lemma: We use induction on $|T|$. Note that $$h(T,Y)=\sum_{\sim}i(T/{\sim},Y)$$ (and similarly for $Z$), where $\sim$ ranges over all equivalence relations on the set $T$, since a map $f:T\to Y$ factors uniquely to give an injection from the quotient of $T$ by the equivalence relation $a\sim b\iff f(a)=f(b)$. By induction on $|T|$, we know $i(T/{\sim},Y)=i(T/{\sim},Z)$ whenever $\sim$ is nontrivial, and we also know $h(T,Y)=h(T,Z)$. It follows that $i(T,Y)=i(T,Z)$.


Now suppose $X$ is a nonempty finite space and $Y$ and $Z$ are finite spaces such that $X\times Y\cong X\times Z$. For any finite space $T$, we have $h(T,X\times Y)=h(T,X)h(T,Y)$ and similarly for $Z$. Since $X$ is nonempty, $h(T,X)\neq 0$ for all $T$. It follows that $h(T,Y)=h(T,Z)$ for all $T$, and hence $i(T,Y)=i(T,Z)$ for all $T$ by the Lemma.

Since $i(Y,Y)>0$ (the identity map exists), we have $i(Y,Z)>0$: there exists a continuous injection $f:Y\to Z$. Similarly, there exists a continuous injection $g:Z\to Y$. The composition $gf:Y\to Y$ is a continuous injection, and hence a bijection since $Y$ is finite. Moreover, $(gf)^n=1_Y$ for some positive integer $n$, and it follows that $gf$ is actually a homeomorphism. Similarly, $fg$ is a homeomorphism. This implies that $f$ and $g$ are homeomorphisms, so $Y\cong Z$.

[If I'm not mistaken, this argument in fact works with topological spaces replaced by any category with a faithful "underlying set" functor to Sets which preserves products and such that any object has a quotient by any equivalence relation on its underlying set, as long as you require $X$ to be such that there exists a map from any object to $X$.]


I submitted a paper, Finitary Categories and Isomorphism Theorems, in 1997 to JPAA, but I don't think it ever got published. A category with finite Hom-sets and quite mild factorization properties has the property that if $X$ and $Y$ are objects such that, for any object $Z$, $\hom(Z,X)$ has as many elements as $\hom(Z,Y)$ then $X$ and $Y$ are isomorphic. If in such a category there is at least one map to an object $A$ from any other object, then $A$ is cancellable for products; i.e. $A \times X$ isomorphic to $A \times Y$ implies $X$ isomorphic to $Y$. I used L. Lovasz (1967) Operations with Structures. Acta Mathematica Academiae Scientiarum Hungaricae Tomus 18 (3-4) pp 321-328, and (1971) On the Cancellation Law among Finite Relational Structures. Periodica Mathematica Hungarica Vol 1 (2), pp 145-156.


Edit 2017.05.14 GRP : I have reviewed Lovasz's paper. While the comments below are generally correct, there are some specific details to be addressed, some of which is done in the paper. I am preparing another answer to highlight some of these details from the paper. An amended form of one of the comments is warranted: I believe a categorical version of the argument has been published, and refers to and is not in Lovasz's paper. End Edit 2017.05.14.

Per request, I am bringing some commentary to this answer.

Prior to Eric Wofsey's nice solution, Laszlo Lovasz proved something similar in 1967 in his paper Operations with structures, Acta Math. Acad. Sci. Hungar. 18 (1967), 321-328. (Thanks to Salvo Tringali for looking up the reference.) I first learned about it from my advisor's book Algebras , Lattices, Varieties by McKenzie, McNulty, and Taylor. Phrased in terms of isomorphism types of finite relational structures, one can cancel in that monoid and also kth roots are unique ($A^k$ isomorphic to $B^k$ means $A$ isomorphic to $B$ as finite powers of finite relational structures). For those referring to the textbook and the paper, Salvo mentions that what is used in the proof of the general statement, is Corollary 3 (p. 321) to Theorem 5.23 in that book, which is, in turn, a very special case of Theorem 4.1 in Lovász's paper.

Chapter 5 of that book also has more to say about when cancellation does not occur for other classes of structures. Being a textbook on Universal Algebra, it takes a view that largely is not category theoretic, which I found appealing as a student.

Gerhard "Much Prefers The Structuralist Approach" Paseman, 2017.05.12.