Reference for: "Surjectivity is just injectivity, only one dimension higher."
When I took a first course in differential topology, I remember reading a blog post(?) that made a comment along the lines of
Surjectivity is injectivity, but one dimension higher.
The content of this statement was that to prove a map $f:M\to N$ was surjective, we could prove that a related map $F:M\times I\to N$ was injective instead, and this would imply surjectivity of $f$. I'm looking to see if anyone knows of the source of this quote, or could explain its content in greater detail.
Solution 1:
It's the other way around: injectivity is surjectivity but one dimension higher. This is a concept that comes up in a variety of homotopy-theoretic arguments. Here is a typical example.
Theorem: Let $n\in\mathbb{N}$ and let $Y$ be a topological space such that $\pi_m(Y,y)$ is trivial for all $m\geq n$ and all basepoints $y\in Y$. Let $X$ be a CW-complex and let $A$ be a subcomplex which contains the $n$-skeleton of $X$. Then the inclusion map $i:A\to X$ induces a bijection $i^*:[X,Y]\to [A,Y]$ where the brackets denote homotopy classes of maps.
Proof sketch: We first prove $i^*$ is surjective. Given a map $f:A\to Y$, we can extend it to $X$ one cell at a time. Extending to each new cell consists of constructing a nullhomotopy of the attaching map of the cell composed with $f$. Since $X$ is $X$ is obtained by attaching cells of dimension $\geq n+1$ to $A$, these attaching maps composed with $f$ always give elements of $\pi_m(Y,y)$ for some $m\geq n$ and some $y\in Y$, and thus are nullhomotopic by hypothesis, so the desired extension exists.
Now for the trick: injectivity is just surjectivity in one dimension higher. Specifically, suppose $f,g:X\to Y$ are homotopic when restricted to $A$. We can then build a map $h:X\times\{0,1\}\cup A\times I\to Y$ which restricts to $f$ and $g$ on $X\times\{0\}$ and $X\times \{1\}$, respectively, and is a homotopy between their restrictions on $A\times I$. To show that $f$ and $g$ are homotopic (and thus actually define the same element of $[X,Y]$), we wish to extend $h$ to $X\times I$. But now note that $X\times I$ can be made into a CW-complex which contains $X\times\{0,1\}\cup A\times I$ as a subcomplex containing its $n$-skeleton. So, by the surjectivity argument applied to $X\times I$ and $X\times\{0,1\}\cup A\times I$, we can extend $h$ to $X\times I$. (In fact, $X\times\{0,1\}\cup A\times I$ actually contains the $(n+1)$-skeleton of $X\times I$, so we only need $\pi_m(Y,y)$ to be trivial for $m\geq n+1$ to get injectivity. In this sense injectivity is "one dimension higher" than surjectivity.)
More generally, suppose you have a map $F:[X,Y]\to Z$ you wish to show is bijective, where $X$ and $Y$ are some spaces and $Z$ is some set. Proving surjectivity amounts to showing that for any element of $Z$, you can construct a map $X\to Y$ which maps to it. Proving injectivity amounts to showing that for any two maps $X\to Y$ which map to the same element of $Z$, you can construct a homotopy between them, i.e. you can extend a certain map $X\times\{0,1\}\to Y$ to a map $X\times I\to Y$. So surjectivity and injectivity have the same flavor (and can in many cases be proved in essentially the same way): they are both about being able to construct certain maps to $Y$. But injectivity is "one dimension higher", since you are constructing a map out of $X\times I$ instead of out of $X$.