Is continuity in topology well-defined? In topology, a function is continuous if inverse of every open set is open. But for the inverse to be well-defined the function should be bijective. For example consider the projection map. It is not injective: $\pi_1(x,y_1)=\pi_1(x,y_2)=x$ with $y_1\neq y_2$. But still it continuous and continuity is proved using the inverse function.
 A: The inverse-image of a set $S$ under a function $f$ is the set $\{x : f(x)\in S\}$.  This exists regardless of whether the inverse function $f^{-1}$ exists.
A: We say that $f:X\rightarrow Y$ is a continuous map if given any open set $U\subset Y$, then $f^{-1}[U]$ is an open set in $X$.  This isn't the inverse map we are using, but a related notion called the "pre-image":
Let $f:X\rightarrow Y$ be a function, and for $B\subset Y$, define the pre-image of $B$ to be
$$f^{-1}[B]=\{x\in X \mid f(x) \in B\}$$
This definition does not require $f$ to be bijective at all, and it is what is being denoted whenever $f^{-1}$ is mentioned.  
(It is also common to use parentheses rather than square brackets, but I've always felt that using the latter is better for being clear about what is meant.)
Worth mentioning is that this induces a map $f^{-1}[\cdot]:\mathcal{P}(Y)\rightarrow \mathcal{P}(X)$ given by sending $B\subset Y$ to $f^{-1}[B]$.  There is also the related map $f[\cdot]:\mathcal{P}(X)\rightarrow \mathcal{P}(Y)$ sending $A\subset X$ to 
$$f[A]:=\{b\in Y \mid \text{there exists $x\in A$ such that $f(x)=b$}\}$$
This also induces a map $f[\cdot]:\mathcal{P}(X)\rightarrow \mathcal{P}(Y)$ sending $A$ to $f[A]$.
However, in general $f^{-1}[\cdot]$ and $f[\cdot]$ are not inverses!  In fact, they are inverses if and only if $f$ has an inverse, for we can consider $f[f^{-1}[\{y\}]]$ and $f^{-1}[f[\{x\}]]$ for $x\in X$ and $y\in Y$.  Injectivity of $f$ provides the identity 
$$f^{-1}[f[A]]=A$$
 and surjectivity of $f$ provides the identity 
$$f[f^{-1}[B]]=B.$$  
