# $U\times W \cong V\times W$ does this imply that $U\cong V$

Let $$U, V,$$ and $$W$$ be smooth manifolds such that $$U\times W$$ is diffeomorphic to $$V\times W$$ does that imply that $$U$$ is diffeomorphic to $$V$$.?

Is this result also true/False in other categories like Groups, Topological spaces?

In the case of smooth manifolds, I think this is true by using rank theorem since by rank theorem the map in local coordinates is of the form $$(x^1,x^2,...,x^n)\to (x^1,x^2,...,x^n)$$ and we can think of $$U$$ as embedded in $$U \times W$$ as $$U\times \{y\}$$ for some $$y\in W$$ fixed. and similarly, think of $$V$$ as embedded in $$V \times W$$ as $$V\times \{y\}$$.

Using rank theorem we can prove that $$U\times \{y\}$$ and $$V\times \{y\}$$ are locally diffeomorphic and use the fact that $$U\times W \cong V\times W$$ to prove that they are in fact diffeomorphic? Is my approach right? If not where does it fail?

For groups, this implies that the statement is true for finite groups.

I have a small question here if $$H\times G \cong K\times G$$ then why cant we quotient on both sides $$\{e\}\times G$$ to get $$H\cong K$$?

Is the result true for topological spaces?

Is there any classification of Categories in which the result is true?

• I have little to no background in differential geometry, but if $W=\emptyset$ is a smooth manifold, then that might give a counterexample. – SmileyCraft Jan 22 '19 at 13:43
• I think it's a good idea to quotient by $\{e\}\times G$, why shouldn't it work ? – Surb Jan 22 '19 at 13:58
• Look up the Whitehead manifold. Its product with $\Bbb R$ is Euclidean space again. There are closed manifold examples as well. – user98602 Jan 22 '19 at 14:22
• Similar questions for groups have been asked here a few times. For example, 1, 2,3, 4. See also 5, on free products. – user1729 Jan 22 '19 at 15:18
• Also, I think this question is interesting, but is far too broad! I can see at least 5 questions: The first question (on smooth manifolds), "Is my approach right?", "Is it true for groups?", "Is it true for topological spaces?", and "Can we classify the categories for which it is true?" – user1729 Jan 22 '19 at 15:27

This doesn't hold if $$U,V,W$$ are allowed to be disconnected. You can create silly counterexamples in the same way as with infinite groups: take $$U = \{0\}$$, $$V = \{1,2\}$$, and $$W = \mathbb{Z}$$. Then $$U\times W \cong \mathbb{Z} \cong V \times W$$ as manifolds, but of course $$U\not\cong V$$. There are obvious higher-dimensional counterexamples of the same nature. (This is similar to $$H = \mathbb{Z}$$, $$K = \mathbb{Z}^2$$, $$G = \mathbb{Z}^{\mathbb{Z}}$$ in groups, or maybe more accurately in free abelian groups.)

As you noted, this question cites this paper which proves the result is true for finite groups, but also that it fails even for $$G = \mathbb{Z}$$ and $$H,K$$ finitely presented! It does also hold for finitely generated (but not necessarily finite) abelian groups by the structure theorem, but it does not hold in general for (abelian) groups (see my example above).

As you can see in the paper cited above, it's a little tricky to get a counterexample for $$G=\mathbb{Z}$$, and I suppose it's also a bit tricky for (closed) connected manifolds, but it looks like Mike Miller knows of one in the comments!

• Sure, but in groups where does the quoting argument fails? – user345777 Jan 22 '19 at 14:13
• The set $\{e\}\times G$ need not be preserved by the isomorphism, by which I mean the isomorphism need not map $\{e_H\}\times G$ to $\{e_K\}\times G$. "Quotienting on both sides by the same thing" only necessarily works when the normal subgroups are in correspondence under the isomorphism. – cs47511 Jan 22 '19 at 14:15
• Sure I get your point – user345777 Jan 22 '19 at 14:25

(Making this CW since the interesting part was done by George Lowther)

First, George Lowther gave a fantastic answer to the smooth manifold question here.

He proves:

There exist connected closed manifolds $$X$$ and $$Y$$ with different homotopy types but for which $$X\times S^1$$ is diffeomorphic to $$Y\times S^1$$.

(This example also answers the question in the category of groups: the fundamental groups of $$X$$ and $$Y$$ are not isomorphic, but $$\pi_1(X)\times \mathbb{Z}\cong \pi_1(Y)\times \mathbb{Z}$$.)

So, in short, NO, you cannot conclude from $$U\times W\cong V\times W$$ that $$U\cong W$$. The problem with your argument is a diffeomorphism from $$U\times W$$ to $$V\times W$$ does not have to map a subset of the form $$U\times \{w\}$$ to a subset of some $$V\times \{w'\}$$, so there is no way to take a diffeomorphism of the larger spaces and reduce it to a diffeomorhpism of the smaller spaces.

The answer is no, but sometimes is yes.

You should see something about the Krull-Remak-Smith theorem in modules and groups and this question on topological spaces.