Algebraic Topology in Simple Terms

I just wanted to clarify a few basic concepts in algebraic topology. Suppose one space is my room ($\text{Room} \ A$). Suppose the other space is another room in my house ($\text{Room} \ B$). So consider the following mappings:

$$f: \text{Room} \ A \to \text{Room} \ B$$ and $$g: \text{Room} \ A \to \text{Room} \ B$$

where $f$ maps points in Room A to the 8 corners of Room B depending on some rules. Also $g$ maps points in Room A to the center of the faces of Room B depending on some rules.

Is the relationship between these two mappings basically what a homotopy is? That is, is it a function $h: f \to g$ (a function between functions)? Suppose we had the following: $$f: \text{Room} \ A \to \text{Room} \ B$$ and $$i: \text{Room} \ A \to \text{Room} \ C$$ where $i$ maps points in Room A to the floor of Room C depending on some rules.

Could we consider the relationship between $f$ and $i$? Would this be a homotopy?

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Hi Robb. To specify a mapping, you need more than the domain and range, but also an assignment of every point in the domain to a point in the range. That is, you haven't actually specified a map from Room A to Room B! –  Grumpy Parsnip Oct 28 '11 at 1:59

There seems to be a little bit of confusion as to what a homotopy is. I like your analogy that a homotopy is a "function between functions" in the sense that a homotopy could be thought of as a transformation (i.e. function) from one function to another. So, let's look at the formal definition for a second. Suppose that we two spaces $X,Y$ (make them as nice as you're comfortable with) and two mappings (continuous as always) $f,g:X\to Y$. So, what does it mean intuitively that $f$ and $g$ should be homotopic? It should mean that that we can get $f$ from $g$ by nicely moving it around. More formally it means that there exists a map $F:I\times X\to Y$ ($I$ is just the unit closed interval) with the property that $F(0,x)=f(x)$ and $F(1,x)=g(x)$. Intuitively what this $F$ represents is the "function between functions" that you mentioned. Namely, for each fixed $t\in I$ we have a natural function $f_t:X\to Y$ given by $f_t(x)=F(t,x)$. Thus, you can imagine $F$ as moving $f$ to $g$ where, we took stop-motion photography, the in-between steps are just the functions $f_t$. Moreover, it's important to notice that we are doing this continuously--we aren't ripping $f$ in half, moving it, and then reassembling it into $g$. To test your intuition, convince yourself that any two loops in $\mathbb{R}^2$ (i.e. paths in $\mathbb{R}^2$ which start and stop at the same point) are homomotopic to the constant map (at their base point). Moreover, convince yourself that if you take a loop encompassing the circle ONCE and a loop encompassing the circle TWICE they can't be homotopic since any way of transformation one into the other will involve some kind of non-allowable action (e.g. ripping or attaching).