# Homeomorphism between a mapping space and a product space

Let $$X$$ and $$Y$$ be some (topological) spaces. The mapping space $$X^Y$$ is the set of maps $$f:X \rightarrow Y$$ endowed with the compact-open topology. A map $$g:Z \rightarrow Y$$ induces a map $$g^*:X^Y \rightarrow X^Z$$ by pre-composition. Key propery of mapping spaces is the following: Let $$i: B\rightarrow Y$$ be a cofibration of locally compact Hausdorff spaces, and let $$X$$ be any space. Then the induced map $$i^*:X^Y \rightarrow X^B$$ is a fibration. I want to show two things:

1. That there is a homeomorphism $$X^{\{0,1\}} \cong X \times X$$, where $$\{0,1\}$$ is given the discrete topology.
2. That the map $$\pi_X:X^I \rightarrow X \times X$$ given by $$\pi_X(\gamma)=(\gamma(0),\gamma(1))$$ is a fibration.

As for the first one, since $$\{0,1\}$$ is discrete, it seems to me that the only good candidate for a homeomorphism is the map $$\pi(\gamma) = (\gamma(0),\gamma(1))$$. We could also consider $$\pi(\gamma) = (\gamma(0),\gamma(0))$$, or $$\pi(\gamma) = (\gamma(1),\gamma(1))$$ but not these are clearly not surjective, right? Is this correct? I am not sure how to show that $$\pi(\gamma)$$ is a continuous and open mapping.

As for the second one, I start with the inclusion map $$i:\{0,1\} \rightarrow [0,1]$$. Since $$\{1,0\}$$ is a subset of $$[0,1]$$, $$i$$ is a cofibration. Moreover, both $$\{1,0\}$$ and $$[0,1]$$ are locally compact Hausdorff spaces. The induced map $$i^{*}:X^{[0,1]} \rightarrow X^{\{0,1\}}$$ is than a fibration. Then I should probably use $$1.$$ to show that $$\pi_X$$ is a fibration, but I'm not sure how to get this one. Any help will be much appreciated.

• As for $1)$, yes these are clearly surjecive: any map out of a discrete space is continuous. In general $X^A\cong\prod_{a\in A}X$ for any finite discrete space $A$. Just use the standard subbase of the compact open topology (a subset of a discrete space is compact if and only if it is fnite). – Tyrone Nov 28 '19 at 12:55

## 1 Answer

You are not explicit whether "compact" includes Hausdorff, so it is not absolutely clear what the compact-open topology is. Let us assume that the Hausdorff property is not included.

You know that if $$g : Z \to Y$$ is continuous, then $$g^* : X^Y \to X^Z, f \mapsto f \circ g$$, is continuous (If "compact" includes Hausdorff, then we must additionally require that $$Y$$ is Hausdorff to assure that $$g^*$$ is continuous. In the context of your question this is satisfied.)

1. Your map $$\pi$$ is a homeomorphism. Since $$\{0,1\}$$ is discrete, all functions $$f : \{0,1\} \to X$$ are continuous which shows that $$\pi$$ is bijective. A subbasis $$\Sigma$$ for $$X^{\{0,1\}}$$ is given by the sets $$\langle K, U \rangle = \{ f \in X^{\{0,1\}} \mid f(K) \subset U \}$$ where $$K \subset \{0,1\}$$ is compact and $$U \subset X$$ is open. Note that all $$K \subset \{0,1\}$$ are compact. We have $$\langle \emptyset, U \rangle = X^{\{0,1\}}$$ and $$\langle \{0,1\}, U \rangle = \langle \{0\}, U \rangle \cap \langle \{1\}, U \rangle$$, thus the set $$\Sigma' = \{ \langle \{i\}, U \rangle \mid i = 0,1 , U \subset X \text{ open } \}$$ is a subbasis for $$X^{\{0,1\}}$$. But $$\pi(\langle \{0\}, U \rangle = U \times X$$ and $$\pi(\langle \{1\}, U \rangle = X \times U$$, and these sets form a subbasis for the product topology on $$X \times X$$. Thus $$\pi$$ is continuous and open, hence a homeomorphism.

2. The inclusion $$i : \{0,1\} \to I$$ is a cofibration. The reason is not that $$\{0,1\}$$ is a subset of $$I$$, but is true because it is closed subset and $$I \times \{0\} \cup \{0,1\} \times I$$ is a retract of $$I \times I$$. You know that $$i^* : X^I \to X^{\{0,1\}}$$ is a fibration. Hence also $$\pi_X = \pi \circ i^*$$ is one simply because $$\pi$$ is a homeomorphism.

• thank you for your answer! And your assumption was correct, Hausdorff property is not included. – billy192 Nov 28 '19 at 13:40