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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.

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    $\begingroup$ 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). $\endgroup$ – Tyrone Nov 28 '19 at 12:55
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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.

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

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