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In Lee's Introduction to Topological Manifolds (first edition) he states the following:

Suppose $A \subset X$ is a subspace. For any topological space $Y$, a map $f: Y \rightarrow A$ is continuous if and only if the following composite map from $Y$ to $X$ is continuous:

$$Y \stackrel{f}{\longrightarrow} A \stackrel{\iota_{A}}{\hookrightarrow} X$$

Proof. Directly from the definitions of continuity and the subspace topology,

$$f : Y \rightarrow A \text{ is continuous}$$

$$\iff \text{for all } U \underset{\text{open}}{\subset} A, f^{-1}(U) \underset{\text{open}}{\subset} Y $$

$$\iff \text{for all } V \underset{\text{open}}{\subset} X, f^{-1}(V \cap A) \underset{\text{open}}{\subset} Y $$

$$*\iff \text{for all } V \underset{\text{open}}{\subset} X, (\iota_{A} \circ f)^{-1}(V) \underset{\text{open}}{\subset} Y $$

$$\iff \iota_{A} \circ f : Y \rightarrow X \text{ is continuous} $$

I get everything before the *. How do I interpret $(\iota_{A} \circ f)^{-1}(V)$?

There are some other posts about the same proof or something similar here and here, but I am specifically interested in understanding the inverse step I just mentioned.

Thanks

matheus
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1 Answers1

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Hint: The following always holds $${(\iota_{A} \circ f)}^{-1}(V) = f^{-1}(\iota_{A}^{-1}(V)) = f^{-1}(V \cap A).$$

One should be able to verify these using elementary set theory.

Duong Ngo
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