Does $A \times B \cong A \times C$ imply $B \cong C$ in arbitrary category? So the question is whether $A \times B \cong A \times C$ implies $B \cong C$ in arbitrary category.
It is true, I suppose, in all of the categories I could think of, so I got interested whether this holds in all categories and this is where I struggled to find proof on my own.
The problem is, proving this statement in particular categories I initially worked within involved using their properties, i.e. them being concrete, and even the specific construction of a product. When I tried proving this using purely categorical definition of a product, I stumbled upon a problem that I only have projections to $A$ and $B$ and $C$, so I was not able to construct any morphism from $B$ to $C$, leave alone isomorphism.
It seems to me that from the universal property of categorical product it somehow follows that all the morphisms from $B$ correspond to morphisms from $C$, thus it may be possible to prove the desired statement using Yoneda's lemma, but I am not very familiar with category theory so I don't know how.
Or is there some elementary proof which I am just too blind to see? Is the statement I want to prove even true in general? I would be really glad to finally figure it out.
I am grateful in advance for any answer or any hints as to how to get to proving or disproving this statement.
 A: Claim: In a category $\cal C$, for all $A,B,C$, one has $A\times B \cong A\times C \Rightarrow B\cong C$.

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*A counterexample in $\bf Set$: take two finite sets $B,C$ and $A$ uncountably infinite.

*A counterexample in abelian groups: take $A=\mathbb{Z}^{(\mathbb N)}$ (direct sum of an infinite number of $\mathbb Z$), $B=0, C=\mathbb Z$: $\bigoplus_{n=0}^\infty \mathbb Z = \bigoplus_{n=0}^\infty \mathbb Z \oplus \mathbb Z = \bigoplus_{n=0}^\infty \mathbb Z \oplus (0)$.

*A counterexample in topological spaces: take $A=[0,1)$ (real interval, open on one side), $B=(0,1), C=[0,1]$.

*A counterexample in a poset, regarded as a category: take as poset the powerset of any nonempty set $X$; $A$ any proper subset (i.e., $A^c$ is not empty or the whole $X$), $B=\varnothing, C=A^c$ (the complement of $A$): $A\cap A^c = \varnothing = A\cap \varnothing$.

At a very informal level, the validity of your claim is bounded to the assumption that the objects of $\cal C$ are in some sense "finite". There has been discussion on this topic in various other threads on MO and MSE.
