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Let $f:X\rightarrow Y$ be a continuous between topological spaces. Show that the restriction of $f$ to any subspace $A \subseteq X$ is continuous.

I am studying for an exam and the answer to this just says that the solution should be obvious from the definition of continuity, so I wanted to check if this is the right answer:

Suppose Y is open, then since $f$ is continous, X is also open. Then, since $A\subseteq$X, A is also open.

So $f:A\rightarrow Y$ is continuous since the preimage of Y, an open set, is also open.

Nicky
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  • I am afraid your notation is a little bit messed up. – Hui Yu Apr 12 '13 at 15:19
  • $@$Nicky: Your answer is not correct. Not even close: if I were grading it, I would give it zero credit (and I am a relatively generous grader even by contemporary American standards). Further, it exhibits some fundamental misconceptions and errors. I suggest that you show this question to your instructor as soon as possible and get more guidance on the basics. – Pete L. Clark Apr 12 '13 at 16:11

2 Answers2

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One possible definition of continuity is the following:

"A function $f:X\rightarrow Y$, where $X$ and $Y$ are topological spaces, is said to be continuous if for every open subset of $Y$, it's preimage by $f$ is open in $X$, i.e., if $V\subseteq Y$ is open (in $Y$), then $f^{-1}(V)$ is open (in $X$)."

In your argument, you only proved that the preimage of a specific open set in $Y$ ($Y$ itself) is open in $A$, which is not sufficient.

Now, let $f:X\rightarrow Y$ be a continuous function and $A\subseteq X$. In the definition of subspace, a subset $V\subseteq A$ is open in $A$ iff there exists $U\subseteq X$ open in $X$ such that $V=U\cap A$.

Consider the restriction $f|_A:x\in A\mapsto f(x)\in Y$, and let $B$ be an open subset of $Y$. It should be clear that $f|_A^{-1}(B)=f^{-1}(B)\cap A$. Since $f$ is continuous from $X$ to $Y$ and $B$ is open in $Y$ then $f^{-1}(B)$ is open in $X$, hence $f|_A^{-1}(B)$ is the intersection of $A$ with an open subset of $X$, and is therefore open in $A$ (for every open subset of $B$).

We then conclude that $f|_A$ is a continuous function from $A$ to $Y$.

Luiz Cordeiro
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Your reasoning is not quite right. The proof goes something like: Let $Y^\prime\subseteq Y$ be open, then by continuity of $f$, $f^{-1}(Y^\prime)\subseteq X$ is open. Hence $f|_A^{-1}(Y^\prime) = f^{-1}(Y^\prime)\cap A$ is open in $A$ by the definition of the subspace topology.

Thus, since the preimage of any open set under $f|_A$ is open, $f|_A$ is continuous.

Abel
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