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I'm studying the book Topology, Geometry, and Gauge Fields by Gregory L Naber. This is for self study.

I'm trying to prove the first part of Lemma 1.1.2

Let $Y$ be a subspace of $Y'$. If $f:X \to Y'$ is a continuous map with $f(X) \subseteq Y$, then, regarded as a map into $Y$, $f:X \to Y$ is continuous.

My attempt is as follows:

To show that $f:X \to Y$ is continuous we need to show that if $U$ is open in $Y$ then $f^{-1}(U)$ is open in $X$. We know that $f:X \to Y'$ is continuous, so for any open set $U'$ of $Y'$ it follows that $f^{-1}(U')$ is open in $X$. Note since $Y$ is a subspace of $Y'$ then it has the relative topology $T = \{ Y \cap U' : U' \in T'\}$, where T' is the topology of $Y'$. So if $U$ is open in $Y$ then $U = Y \cap U'$ for some open subset $U'$ of $Y'$.

So now we have $f^{-1}(U) = f^{-1}(Y \cap U')$ and it is at this point where I'm having trouble moving forward. If this function were injective then I could simply write $f^{-1}(Y \cap U') = f^{-1}(Y) \cap f^{-1}(U') = X \cap f^{-1}(U')$ which is open. But I'm not given that it is injective so somehow I need to use the fact that $f(X) \subseteq Y$. I feel like I need to concentrate on $f^{-1}(Y \cap U')$ in conjunction with $f(X) \subseteq Y$.

Am I on the right track here? I'm just not sure of how to move forward. Thanks

a_a
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1 Answers1

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Because $f(X) \subseteq Y$, $f^{-1}(W) = f^{-1}(Y \cap W)$ for any set $W \subseteq Y'$. (Proof: If $x \in X$ is such that $f(x) \in W$, then $f(x) \in Y \cap W$. Conversely, if $f(x) \in Y \cap W$, then $f(x) \in W$.)

So, in your proof, we have $f^{-1}(U) = f^{-1}(Y \cap U') = f^{-1}(U')$, which is open.

Nick Matteo
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