# $X_1\cong X_2$ and $Y_1 \cong Y_2$ implies $X_1 \times Y_1 \cong X_2 \times Y_2$

To be shown is that if $X_1 \cong X_2$ and $Y_1 \cong Y_2$ then $X_1 \times Y_1 \cong X_2 \times Y_2$.

I know that one can easily write down a homeomorphism showing this, but I wanted to show it by using the fact that two final objects in the same category are isomorphic. However, my proof was considered incomplete by the rater, and I don't see why. Here my proof:

We know that $X_1 \times Y_1$ and $X_2 \times Y_2$ together with their projection maps are final objects in the corresponding catgory $\mathscr{C}_1$, $\mathscr{C}_2$ (the objects are topological spaces with continuous maps to $X_1$,$Y_1$ and $X_2$,$Y_2$, respectively).

Let $(A, f, g) \in Ob(\mathscr{C}_1)$. There is precisely one morphism $h_1: (A, f, g)\to (X_1 \times Y_1, p_{X_1}, p_{Y_1})$. Since we have homeomorphisms $j: X_1 \to X_2$ and $k: Y_1\to Y_2$, we conclude that $h_1$ is also a morphism from $(A, j\circ f, k\circ g) \in Ob(\mathscr{C}_2)$ to $(X_1 \times Y_1, j \circ p_{X_1}, k \circ p_{Y_1})\in Ob(\mathscr{C}_2)$.

Suppose $h_2$ is another morphism. In particular, $j \circ p_{X_1} \circ h_2$= $j\circ f$. But then $j^{-1}\circ j \circ p_{X_1} \circ h_2=j^{-1} \circ j\circ f$, thus $p_{X_1} \circ h_2= f$. Likewise, we obtain $p_{Y_1} \circ h_2= g$. Therefore, $h_2$ is also a morphism $(A, f, g)\to (X_1 \times Y_1, p_{X_1}, p_{Y_1})$. Since $(X_1 \times Y_1, p_{X_1}, p_{Y_1})\in Ob(\mathscr{C}_1)$ is a final object, it follows $h_2=h_1$.

Now we observe that all objects of $\mathscr{C}_2$ have the form $(A, j\circ f, k\circ g)$: If there are continuous maps from $A$ to $X_2$ and $Y_2$, then the composition of $j^{-1}$ and $k^{-1}$ with these maps are continuous maps from $A$ to $X_1$ and $Y_1$. Therefore, all the continuous maps from $A$ to $X_2$ and $Y_2$ can be obtained from the continuous maps from $A$ to $X_1$ and $Y_1$ (and vice versa). Since $A$ was arbitrary, we conclude that $(X_1 \times Y_1, j \circ p_{X_1}, k \circ p_{Y_1})\in Ob(\mathscr{C}_2)$ is a final object, which implies $X_1 \times Y_1 \cong X_2 \times Y_2$.

• Please use paragraphs. – Cameron Williams Nov 28 '15 at 20:22
• Also, usually people use $\simeq$ for homotopy equivalent and $\cong$ for homeomorphic. – Daniel Robert-Nicoud Nov 28 '15 at 20:25
• The proof is fine IMHO, and works for isomorphisms in any category (where the product exists at least for the objects in question) – Hagen von Eitzen Nov 28 '15 at 20:35
• I've added paragraphs now. My raters comment to the last Paragraph was: "What you really need here is $X_2\times Y_2 \overset{h_1}{\to} X_1\times Y_1 \overset{\varphi}{\to} X_2\times Y_2$ so that $\varphi \circ h_1 = \mathrm{id}$." Can you make sense of it? – Fabian Nov 28 '15 at 21:05
• Just use the product of the homotopies and throrw in a homotopy. – user60589 Nov 28 '15 at 21:29