Topology and quotient spaces involving inverse function 
Let $X$ be a set and let ($Y, \tau)$ be a topological space. Let $g : X \to Y$ be a given map. Define $\tau'=\{U \subset X : U = g ^{−1} (V ) \text { for some } V \in \tau\}$. Which of the following statements are true? 
a. $\tau^{\prime}$ defines a topology on $X$. 
b. $\tau'$ defines a topology on $X$ only if $g$ is onto. 
c. Let $g$ be onto. Define the equivalence relation $x \sim y$ if, and only if, $g(x) = g(y)$. Then the quotient space of $X$ with respect to this relation, with the topology inherited from τ', is homeomorphic to (Y, τ).

I think a. is true, but am unable to show it why especially because the map $g$ is not given to be continuous. Whether onto implies continuity is unable to be proved by me. As for third one, the claim seems true, but the exact line of reasoning is quite elusive to me. Any help. Thanks beforhand.
 A: (a) has nothing to do with $g$ being continuous. Continuity doesn't even have any meaning here because $X$ doesn't have a topology defined on it.
To see it's a topology, just blindly follow the axioms using the definition of open set provided: if $U_i\in\tau'$ for $i$ in some index set $I$, then $U_i=g^{-1}(V_i)$ for $V_i\in\tau$. Then
$$\cup_i U_i=\cup_i g^{-1}(V_i)=g^{-1}(\cup_i V_i)$$
which is in $\tau'$ because $\cup_i V_i\in\tau$ by the definition of a topology. Use the same approach to show that $\tau'$ is closed under finite intersections, and that $\varnothing,X\in\tau'$.
Since we've now shown (a) is true, (b) is certainly false.
I'll tell you as much that (c) is true, but you should try to prove it yourself. A hint is that there is a very obvious map from $X/\mathord\sim$ to $Y$.
A: (a) Is true. The topology $\tau'$ is just the minimal topology on $X$ that makes $g$ continuous. For this we need all sets of the form $g^{-1}[O]$ for $O \in \tau$ to be in the topology. The facts that $\tau$ is a topology and that $g^{-1}$ preserves unions and intersections, imply that these sets already form a topology.
This idea is used in the subspace topology as well:if $X$ would be a subset of $Y$ and $g(x) = x$ would be the standard embedding, sets of the form $g^{-1}[O] = O \cap X$ wouldform the subspace topology on $X$.
This plus the fact that surjectivity of $g$ is irrelevant in the proof of (a), shows (b) is false.
As to c) we introduce some notation, $X / \sim = \{[x]; x \in X\}$ is the quotient space (the set of classes) w.r.t. $\sim$ defined as $x \sim y$ iff $g(x) = g(y)$ (which one easily checks to be an equivalence relation by properties of $=$), and $q(x) = [x]$ is the standard quotient map that sends $x$ to its class, and $O \subseteq X /\sim$ is open iff $q^{-1}[O]$ is open in $X$ (so in $\tau'$).
To define a homeomorphism between $X/\sim$ and $Y$ when $g$ is surjective, the idea is obvious: define $\hat{g}([x]) = g(x)$, for $[x] \in X/\sim$. 
This is well-defined because if we choose another representative for $[x]$, $[x] = [x']$ iff $x \sim x'$ iff $g(x) = g(x')$, the result of the definition is the same. This definition makes a diagram commutative: $\hat{g} \circ q = g$ by definition.
$\hat{g}$ is injective: suppose $\hat{g}([x]) = \hat{g}([x'])$ which means by the definition that $g(x) = g(x')$, and this says that $x \sim x'$ so $[x] = [x']$, as required.
$\hat{g}$ is surjective when $g$ is: if $y \in Y$ we pick $x \in X$ with $g(x) = y$, but then $\hat{g}([x]) = g(x) = y$ and $y$ is in the image of $\hat{g}$.
$\hat{g}$ is continuous: suppose $O \subseteq Y$ is open. Then $\hat{g}^{-1}[O]$ is open in $X/\sim$ iff (definition of quotient topology) $q^{-1}[\hat{g}^{-1}[O]]$ is open in $\tau'$, but $x \in q^{-1}[\hat{g}^{-1}[O]]$ iff $q(x) = [x] \in \hat{g}^{-1}[O]$ iff $\hat{g}([x]) = g(x) \in O$ iff $x \in g^{-1}[O]$, so  $q^{-1}[\hat{g}^{-1}[O]]  =g^{-1}[O]$ which is in $\tau'$ by definition.
$\hat{g}$ is open: suppose $O \subseteq X/\sim$ is open. This means by definition that $q^{-1}[O] \in \tau'$, so $q^{-1}[O] = g^{-1}[U]$ for some $U \in \tau$.
Claim: $\hat{g}[O] = U$ (which is then open, as required). To see the claim: Suppose $y \in \hat{g}[O]$, then $y = \hat{g}([x]) = g(x)$ for some $[x] \in O$. But  $[x] = q(x)$, so $x \in q^{-1}[O] = g^{-1}[U]$, but then $y  =g(x) \in U$. And if $y \in U$, pick $x$ with $g(x) = y$. ($g$ is surjective), so $x \in g^{-1}[U]$, and so $x \in q^{-1}[O]$, which means that $[x] \in O$ and so $y = g(x) = \hat{g}([x]) \in \hat{g}[O]$, showing the second inclusion.
Quite straightforward, though a bit tedious checking..
