show that $f$ is continuous if and only if $\tau'$ is finer than $\tau$.

Question

let $A$ and $A'$ denote a single set in two topologies $\tau$ and $\tau '$ respectively on set $X$.

let $f:A'\to A$ be identity map.

show that $f$ is continuous if and only if $\tau'$ is finer than $\tau$.

let $U$ any open set in $(X,\tau)$

want to show $U$ is open in $(X,\tau')$ assuming $f$ is continous.

it mean $U\cap A$ is open in $(A,\tau_A)$ and hence $f^{-1}(U\cap A)=U\cap A\ $ is open in $(A',\tau_{A'})$

From this how will we show that $U$ is open in $(X,\tau')$

any hint??thanks in advanced

Answer 1

I suppose that your problem is this:

Let $X$ be a set and consider on $X$ two topologies $\tau$ and $\tau'$. Let $f\colon(X,\tau')\longrightarrow(X,\tau)$ be the identity function. Show that $f$ is continuous if and only if $\tau'$ is finer than $\tau$.

If $f$ is continuous and $A$ is an element of $\tau$, then, by definition, $A$ is an open subset of $(X,\tau)$. Therefore, $f^{-1}(A)$ is an open subset of $(X,\tau')$. In other words, $A\in\tau'$. This proves that $\tau'$ is finer than $\tau$.

And if $\tau'$ is finer than $\tau$, then, if $A\in\tau$, $f^{-1}(A)(=A)\in\tau'$. Therefore, $f$ is continuous.