Homeomorphism of Coproduct

Question

Consider the space $C$ and maps $f_A: A \to C$ and $f_B: B \to C$. Assume that there exists a unique continuous function $G: C \to T$ such that $g_A= G \circ f_A$ and $g_B= G \circ f_B$ for any space T and pair of continuous functions $g_A: A \to T$ and $g_B: B \to T$. I wish to show that $C$ and $A \sqcup B$ are homeomorphic.

My idea is to leverage the hypothesis after taking $T = A \sqcup B$. Ultimately, I would like to identify a unique continuous function $F: A \sqcup B \to C$. However, I am unsure of how to proceed.

When proving a similar result for $A \times B$, I leveraged the fact that the projection maps are continuous. I do not know if something similar can be applied here.

Answer 1

In the comments you point out that you are not assuming that $f_A$ and $f_B$ are continuous (something both I and Mark Saving assumed in our responses).

If you do not assume they are continuous, then in general the conclusion you want is false.

We know that any map $f\colon X\to Y$ between topological spaces in which $X$ has the discrete topology is continuous. And a one-to-one map $g\colon X\to Y$ in which $Y$ has the discrete topology is continuous if and only if $X$ has the discrete topology as well.

Thus, if we endow the disjoint union $A\amalg B$ of the underlying sets of $A$ and $B$ with the discrete topology, and let $f_A$ and $f_B$ be the canonical set-theoretic embeddings, then these maps are not continuous unless both $A$ and $B$ have themselves the discrete topology on them.

However, it is definitely the case that given any topological space $T$ and any continuous maps $g_A\colon A\to T$ and $g_B\colon B\to T$, there exists a unique continuous function $G\colon A\amalg B\to T$ such that $g_A=G\circ f_A$ and $g_B=G\circ f_B$, because there is a unique set-theoretic function with those properties, and since $A\amalg B$ has the discrete topology then any function with domain $A\amalg B$ is continuous, in particular $G$.

But $A\amalg B$ is homeomorphic to $A\sqcup B$ if and only if $A$ and $B$ both have the discrete topology (and in that case, we do have that $f_A$ and $f_B$ are continuous). So the conclusion you want need not hold if $f_A$ and $f_B$ are not assumed to be continuous.

You can construct other examples by simly taking any topology on $A\sqcup B$ which his strictly finer than the disjoint topology.

So we must assume that $f_A$ and $f_B$ are continuous for the desired conclusion to hold.

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Under that assumption...

This is a standard argument for objects defined via universal properties.

Presumably you know that you have continuous maps $\iota_A\colon A\to A\sqcup B$ and $\iota_B\colon B\to A\sqcup B$, the canonical inclusions, which have the property that for any topological space $X$ and any continuous maps $g_A\colon A\to X$ and $g_B\colon B\to X$, you have a unique continuous map $G\colon A\sqcup B\to X$ such that $g_A=G\circ \iota_A$ and $g_B=G\circ \iota_B$; namely, $G$ is given by defining it as $g_A$ on $A$ and as $g_B$ on $B$, once you identify $A$ and $B$ as subsets of $A\sqcup B$...

So then, $A\sqcup B$ has the same universal property as your putative $C$. So this sets up the argument to show they are isomorphic/homeomorphic:

1. Because we have continuous embeddings $\iota_A\colon A\to $ and $\iota_B\colon B\to T$, there exists a unique continuous function $\Phi\colon C\to A\sqcup B$ such that $\iota_A=\Phi\circ f_A$ and $\iota_B=\Phi \circ f_B$.

2. Because we have continuous embedding $f_A$ and $f_B$ into $C$, the property of $A\sqcup B$ yields a unique continuous $\Psi\colon A\sqcup B\to C$ such that $f_A=\Psi\circ \iota_A$ and $f_B=\Psi\circ\iota_B$.

3. Now consider the maps $f_A$ and $f_B$ into $C$. We should have a unique continuous function $\Theta\colon C\to C$ such that $f_A=\Theta\circ f_A$ and $f_B=\Theta\circ f_B$. Since $\mathrm{Id}_C$ certainly fits the bill, then $\Theta=\mathrm{Id}_C$ is the only map what will "work". However, consider now what happens with $\Psi\circ\Phi$. We have: $$\begin{align*} (\Psi\circ\Phi)\circ f_A &= \Psi\circ(\Phi\circ f_A) = \Psi\circ\iota_A = f_A\\ (\Psi\circ\Phi)\circ f_B &= \Psi\circ(\Phi\circ f_B) = \Psi\circ\iota_B = f_B. \end{align*}$$ So $\Psi\circ\Phi$ also "works". By the uniqueness clause, we conclude that $\Psi\circ\Phi = \Theta = \mathrm{Id}_C$.

Now try to do something similar to show that $\Phi\circ\Psi=\mathrm{Id}_{A\sqcup B}$ and conclude that you have homeomorphisms.

Answer 2

This is a special case of the uniqueness of coproducts up to unique isomorphism. Note that in the context of topological spaces, isomorphism and homeomorphism are synonymous.

More formally, we say that $(C, f_A : A \to C, f_B : B \to C)$ is a coproduct of $A$ and $B$ if and only if for all $g_A : A \to T$ and $g_B : B \to T$, there exists a unique $g_C : C \to T$ such that $g_C \circ f_A = g_A$ and $g_C \circ f_B = g_B$.

This definition makes sense in any category.

One can then prove that coproducts are unique up to unique isomorphism. That is, given two coproducts $(C, f_A, f_B)$ and $(D, g_A, g_B)$ of $A$ and $B$, there exists a unique isomorphism $i : C \to D$ such that $g_A = i \circ f_A$, $g_B = i \circ f_B$.

To prove this, we know that there is a unique $i : C \to D$ such that $i \circ f_A = g_A$ and $i \circ f_B = g_B$. This is because $C$ is a coproduct. Now, we just need to prove that $i$ is actually an isomorphism.

We can also take the unique $j : D \to C$ such that $j \circ g_A = f_A$ and $j \circ g_B = f_B$ by the fact that $D$ is a coproduct.

Combining these facts, we see that the map $h = i \circ j : D \to D$ satisfies $h \circ g_A = g_A$ and $h \circ g_B = g_B$. Since $D$ is a coproduct, there is exactly one such function $h$. Note that it is also the case that $1_D \circ g_A = g_A$ and $1_D \circ g_B = g_B$, where $1_D : D \to D$ is the identity map. Therefore, $h = 1_D$. That is, $i \circ j = 1_D$.

The same argument shows that $j \circ i = 1_C$. Therefore, $i$ and $j$ are inverse isomorphisms.

So all we have to do to show that $C$ and $A \sqcup B$ are isomorphic (homeomorphic) is to show that $A \sqcup B$ is the coproduct of $A$ and $B$.

We know that there are canonical inclusions $i_A : A \to A \sqcup B$ and $i_B : B \to A \sqcup B$ such that $(A \sqcup B, i_A, i_B)$ is the coproduct of $A$ and $B$ in the category of sets. We also know that a subset $S \subseteq A \sqcup B$ is open if and only if $i_A^{-1}(S)$ and $i_B^{-1}(S)$ are both open. This guarantees that $i_A$ and $i_B$ are continuous.

Now given continuous functions $g_A : A \to T$, $g_B : A \to T$, consider the unique function (not necessarily continuous) $h : A \sqcup B \to T$ such that $h \circ i_A = g_A$, $h \circ i_B = g_B$, which exists because $(A \sqcup B, i_A, i_B)$ is a coproduct of $A$ and $B$ as sets.

I claim that $h$ is continuous. Indeed, consider an arbitrary open set $U \subseteq T$. Then $i_A^{-1}(h^{-1}(U)) = (i_A \circ h)^{-1}(U) = g_A^{-1}(U)$ is open because $g_A$ is continuous. And similarly, we see that $i_B^{-1}(h^{-1}(U))$ is also open. Therefore, $h^{-1}(U)$ is open.

Thus, we see that $(A \sqcup B, i_A, i_B)$ is a coproduct of $A$ and $B$ in the category of topological spaces, hence homeomorphic to $C$.