1

Let $X$ and $Y$ be topological spaces and $f$ a function of $X$ into $Y$. Show that $f$ is continuous if and only if es continuous as a function of $X$ onto the subspace $f(X)$ of $Y$.

I'm proceding like this:

First, assume that $f$ is continuous of $X$ into $Y$, let $f(X) \cap A$ be an open set of $f(X)$, where $A$ is open in $Y$. So $f^{-1} (f(X)\cap A)=f^{-1}(f(X)) \cap f^{-1}(A)$, $ f^{-1}(A)$ is open because $f$ is continuous , and $X=f^{-1}(f(X))$ because $f$ is a function of $X$ into $Y$, so $f^{-1} (f(X)\cap A)$ is open.

¿Am I proceeding right? , and ¿How can I prove the reverse?, I know that it's a easy problem but I'm stuck.

Giotaker
  • 226

3 Answers3

1

Yes, you are proceeding right. To show the backwards direction, let $A$ be open in $Y$. Then, $A \cap f(X)$ is open in $f(X)$ and so $f^{-1}(A \cap f(X))$ is open in $X$. But $f^{-1}(A \cap f(X)) = f^{-1}(A)$.

0

What you have so far looks good. For the converse, suppose that $f:X\to f(X)$ is a continuous function. Let $A\subset Y$ be open. Show that $f^{-1}(A)=f^{-1}(A\cap f(X))$.

Plutoro
  • 22,730
  • 3
  • 41
  • 71
0

If you like to be pedantic, one could say this is an immediate consequence to the characterisation of continuity of maps into a space with the initial topology. I.e. if we have $f: X \rightarrow Y$ and $i: f[X] \rightarrow Y$ is the inclusion map, and $\tilde{f}: X \rightarrow f[X]$ is the image restriction of $f$, so $\tilde{f}(x) = f(x)$ for all $x \in X$, then $i \circ \tilde{f} = f$ and $f$ is continuous iff $\tilde{f}$ is, by the initial topology fact ($f[X]$ has the initial topology w.r.t. $i$, being a subspace topology).

Community
  • 1
Henno Brandsma
  • 242,131