Show that $T$ is a topology on $X$. Let $f:X\rightarrow Y$ be a map and let $T'$ be a topology on $Y$.
Show that $$T=\{U\subseteq X \mid \exists V\in T' \text{ with }U=f^{-1}(V)\} $$ is a topology on $X$.
$T'$ is the coarsest topology on $X$ such that the map $f:(X, T) \rightarrow (Y, T') $ is continuous.
$$$$
We have to show:
1.The set $X$ and the empty set are elements of $T$ .


*Any union of elements of $T$ belongs to $T$.


*Any finite intersection of elements of $T$ belongs to $T$.
$$$$
Toshow these axioms, Ihave done the following:

*

*It holds that $X\subseteq X$ and $f(X)\in T'$ where $T'$ is a topology, and so we get $X\in T$.
We also have that $\emptyset\subseteq X$ and $f(\emptyset)\in T'$ where $T'$ is a topology, and so we get $\emptyset\in T$.


*We consider the union $O = \bigcup\limits_{\alpha \in I} U_{\alpha}$.
This set is in $T$ iff theimage under $f$ is in $T'$.
So we consider $f \left( O \right)$.
The union is well defined under images so we get \begin{equation*}f \left( O \right) = f \left( \bigcup\limits_{\alpha \in I} U_{\alpha} \right) = \bigcup\limits_{\alpha \in I} f \left( U_{\alpha} \right)\end{equation*}
Since each $U_{\alpha}$ is in $T$, the images $f\left( U_{\alpha} \right)$ are in $T'$.
Since $T'$ is a topology, is the union again in $T'$ and so we get that $O$ in $T$.


*The respective argument of 2.for intersection.

Is everything correct?
 A: $\emptyset = f^{-1}[\emptyset]$ and as $\emptyset \in T'$ we have by definition $\emptyset \in T$. Same for $X=f^{-1}[Y]$ and $Y \in T'$.
If $U,V \in T$ so $U = f^{-1}[U']$ for some $U' \in T'$ and $V = f^{-1}[V']$ for some $V' \in T'$, we have $U \cap V =  f^{-1}[U'] \cap f^{-1}[V']= f^{-1}[U' \cap V']$ and $U' \cap V' \in T'$, as $T'$ is closed under finite intersections, so $U \cap V \in T$ as well.
If for all $i \in I, U_i \in T$ so that we have $U'_i \in T'$ such that $f^{-1}[U'_i]=U_i$, then $\bigcup_i U'_i \in T'$ too and as $\bigcup_i U_i = \bigcup_i f^{-1}[U'_i] = f^{-1}[\bigcup_i U'_i]$ we get that $\bigcup_i U_i \in T$, as required.
So you have to write everything as preimages from $T'$, not check the image is in $T'$, because that is how $T$ is defined.
If $U$ is any topology on $X$ that makes $f: (X,U) \to (Y,T')$ continuous then for any $O \in T'$ we must have $f^{-1}[O] \in U$, but by definition, $f^{-1}[O]$ is the form of any set in $T$ so $T \subseteq U$. This shows that $T$ is the smallest (coarsest) topology that makes $f$ continuous.
A: So, we made two corrections, and you are well on your way!
Left is just to prove the intersection and union properties.  But they follow from the general facts:

*

*$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$

*$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$
Note there is another correction here:  you want preimages not forward images, because you're starting with the topology $T'$ on $Y$.
That $T$ is the coarsest topology that makes $f$ continuous is clear, because if you take out even one open set, the definition fails.
