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Let $(X, \mathcal{T})$ and $(Y, \mathcal{S})$ be two topological spaces.

Consider a function $\gamma: X \to Y$ and the corestricted function $\eta: X \to \gamma(X): x \mapsto \gamma(x)$. I want to prove the following:

$\gamma$ is continuous if and only if $\eta$ is continuous (the codomain is equipped with the subspace topology).

Attempt:

Suppose $\gamma$ is continuous. Let $A$ be open in the subspace $\gamma(X)$. Then $A = \gamma(X) \cap B$ where $B \in \mathcal{S}$. Hence

$$\eta^{-1}(A) = \eta^{-1}(\gamma(X) \cap B) = \{x \in X\mid \eta(x) \in \gamma(X)\cap B\}= \{x \in X\mid \gamma(x) \in \gamma(X)\cap B\}$$ $$= \{x \in X \mid \gamma(x) \in B\}= \gamma^{-1}(B)\in \mathcal{T}$$

Conversely, suppose that $\eta$ is continuous. Let $A \in \mathcal{S}$. Then

$$\gamma^{-1}(A) = \gamma^{-1}(A \cap \gamma(X)) = \eta^{-1}(A \cap \gamma(X))$$

and since $A \cap \gamma(X)$ is open in the subspace $\gamma(X)$ of $S$, it follows that $\gamma^{-1}(A) \in \mathcal{T}$.

Is this proof correct?

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    You proved that $\eta^{-1}(A)=\eta^{-1}(B)$, but $\eta$ has not proven to be continuous yet! What should be proved is $\eta^{-1}(A)=\gamma^{-1}(B)$. The proof for $\eta$ continuous $\Rightarrow$ $\gamma$ continuous is correct. – withoutfeather Sep 01 '19 at 16:29
  • @withoutfeather Thanks for your useful remark! I edited my proof. Is it correct now? –  Sep 01 '19 at 16:58

1 Answers1

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The general view is to use the notion of initial topologies and their universal property, see my answer here. Let $e: \gamma[X] \to Y$ be the inclusion map ($e(x)=x$) and then $\gamma[X]$ has the initial topology w.r.t. $e$ (usually known for this special case as the subspace topology). In particular, $e$ is continuous and $$e \circ \gamma = \eta\tag{1}$$

as is easily checked. If $\gamma$ is continuous, so is $\eta$ as the composition of continuous maps. And if $\eta$ is continuous, the universal property of the initial topology (explained in the linked answer) implies $\gamma$ is continuous too. So the asked for equivalence is then almost immediate.

Henno Brandsma
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  • Thanks! I should really revise initial/final topologies! Could you verify if my approach works too? –  Sep 01 '19 at 20:19
  • @user661541 Yes, the proof you gave seems correct to me. But I prefer to have general theory to fall back on. – Henno Brandsma Sep 01 '19 at 20:20
  • Yes, I realise your approach is much shorter. Observing that we can view one function as a composition with the inclusion map and the other made things cery clear. Thanks again. –  Sep 01 '19 at 20:21
  • @user661541 such a category theory view (based on arrows and compositions and "limits") can be enlightening sometimes. – Henno Brandsma Sep 01 '19 at 20:23
  • Yes, I think I will appreciate this even more once I have followed a category theory course. –  Sep 01 '19 at 20:24
  • My university even a course on categorical topology I believe, which I intend to take. –  Sep 01 '19 at 20:25
  • @user661541 That's quite rare. Who teaches it, if I may ask? – Henno Brandsma Sep 01 '19 at 20:26
  • I prefer not to say –  Sep 01 '19 at 20:27