In Munkres' Topology there is a theorem which states that the image of a connected space under a continuous map is connected.
In the proof of this, they let $f: X \rightarrow Y$ be continuous, where $X$ is connected, and prove that $Z = f(X)$ is connected. It says that since the restricted map to $Z$ is also continuous, it suffices to consider $g: X \rightarrow Z$. Supposing that $Z = A \cup B$ is a separation, then $g^{-1}(A)$ and $g^{-1}(B)$ are disjoint nonempty open sets whose union is $X$. And then they form a separation of $X$ which is contradicting the fact that $X$ is connected.
So here is my question: The reason we know that $g^{-1}(A)$ and $g^{-1}(B)$ are nonempty is because $g$ is a surjection (according to the book). But how do we know for sure that $g$ is a surjection? We are told that $f$ is continuous and that $g$ is a restriction to $Z$, but what if $f$ is injective, couldn't $g$ be injective as well?
Best Answer
To say that $Z=f(X)$ means that $Z = \{f(x) \,|\, x \in X\}$. The restriction of the function $f : X \to Y$ to the subset $Z \subset Y$ is the function $g : X \to Z$ defined by $g(x)=f(x)$ for all $x \in X$ (sometimes this is called a "range restriction", as opposed to the usual notion of restriction which might be called a "domain restriction").
If we take any element $z \in Z$, then applying the definition of $Z$ we have $z=f(x)$ for some $x \in X$, and then applying the definition of $g$ we have $z=g(x)$. In other words, for each $z \in Z$ there exists $x \in X$ such that $g(x)=z$. This is exactly what it means for the function $g$ to be surjective.
By the way, $g$ might well be injective, but you asked about surjectivity which is different than (and logically independent from) injectivity.