Let be $\left ( X,\tau \right )$ a path-connected space. Let be $\left ( X,{\tau}' \right )$ a space such that $\left ( X,{\tau}' \right ) \subset \left ( X, \tau \right )$ ($\tau$ is finer than ${\tau}'$). Show that $\left ( X,{\tau}' \right )$ is path-connected.
My attemp:
Let be $a,b \in \left ( X,{\tau}' \right )$. For all $x,y \in \left ( X,\tau \right )$ there is a path $f:\left [ 0,1 \right ] \rightarrow \left ( X,\tau \right )$ from $x$ to $y$, i.e. $f$ is continuous and $f\left ( 0 \right )=x$ and $f\left ( 1 \right )=y$. Particularly for $a,b \in \left ( X,\tau \right )$ there is a path $F:\left [ 0,1 \right ] \rightarrow \left ( X,\tau \right )$.
The identity function $i : \left ( X,\tau \right ) \rightarrow \left ( X,{\tau}' \right )$ is continuous because given $U\in{\tau}'$, $i^{-1}\left ( U \right ) = \left \{ x \in \left ( X,\tau \right )\mid i\left ( x \right )=x\in U \right \}=U$ is open in $\tau$ because ${\tau}' \subset \tau$. Then $i\circ F:\left [ 0,1 \right ] \rightarrow \left ( X,{\tau}' \right )$ is a path from $a$ to $b$.
Therefore $\left ( X,{\tau}' \right )$ is path-connected.
If I'm wrong give me a hint to be right, please. Thanks in advance.
Yes, the idea is essentially that making the topology in the codomain smaller, any continuous function stays continuous. Your proof is correct. Recap:
So let $a,b \in X$. By path continuity of $(X,\tau)$ there is a continuous $p: [0,1] \to (X,\tau)$ with $p(0)=a, p(1)=b$. As $\tau' \subseteq \tau$, that same map regarded as $p: [0,1] \to (X,\tau')$ is also continuous (as $O \in \tau' \to O \in \tau \to p^{-1}[O] \text{ open in } [0,1]$) and so that $p$ is a path from $a$ to $b$ in $(X,\tau')$ as well. As $a,b$ were arbitrary, $(X,\tau')$ is path-connected.