# Is there any analogy between homotopy of spaces and homotopy between maps?

I am a beginner of Algebraic topology course. I just come across the definition of Homotopy

Homotopy of 2 maps:

$$f,g:S\to T$$ are 2 continuous maps is said to called homotopic if there exist continuous map $$H:S\times [0,1]\to T$$ such that $$H(s,0)=f(s)$$ and $$H(s,1)=g(s)$$.

Homotopy of 2 spaces :

Two topological spaces $$X,Y$$ are said to homotopic if there is continous map $$f:S\to T$$ and $$g:T\to S$$ such that $$f\circ g:T\to T$$ is homotopic to Identity on T and $$g\circ f:S\to S$$ is homotopic to Identity on S.

I know that to define second definition we used the first one.Also intuitively for the first defination, if we consider graphs of 2 function then one continuously deformed to another

Is this continuous deformation happen in 2nd definition .But how to interpret that from definition only?

This question arises when I am reading and try to relate both definition

• I wouldn't say that $X$ and $Y$ are homotopic in your situation. I'd say they were homotopy equivalent. – Lord Shark the Unknown Jul 10 at 6:32
• Dear Sir , Is it not possible to continuously deform to get another space? I know that 2 spaces can be completely different but up to some isomorphism can we do? – MathLover Jul 10 at 6:35
• If I am not wrong then you are probably trying to understand the "big-picture" behind "continuous deformation of two topological spaces" and "continuous deformation of two continuous maps". – user 170039 Jul 10 at 6:42
• Unfortunately I am not an expert in this area. But I am sure other people can help you. Just wait a bit. Also I would advise you not to refer others as "Sir" as there are several people in this site who doesn't like to be called as such. – user 170039 Jul 10 at 6:49

You have mentioned graphs of functions: indeed it is quite clear that if $$f\sim g$$ then the graph of $$f$$ $$\subset S\times T$$ is homotopy equivalent to the one of $$g$$, however the converse is far from true, as any two graphs are homotopy equivalent. In fact, any two graphs are homeomorphic : let $$f:S\to T$$ be any continuous map and $$\Gamma_f = \{(s,t)\in S\times T, s\in S, f(s) =t\}$$. Then we have a map $$i:S\to \Gamma_f, s\mapsto (s,f(s))$$ and a map $$\Gamma_f \to S, (s,t)\mapsto s$$ which are both clearly continuous, and inverse to one another.
Then we have the following : given a space $$S$$, a cylinder for $$S$$ is a space $$S_{\wedge I}$$ together with maps $$in_0, in_1 : S\to S_{\wedge I}$$ and a homotopy equivalence $$p:S_{\wedge I}\to S$$ such that $$p\circ in_0 = id_S, p\circ in_1 = id_S$$.
Then, two maps $$f,g: S\to T$$ are homotopy equivalent if and only if there is a cylinder for $$S$$, $$(S_{\wedge I},in_0,in_1,p)$$ and a map $$H:S_{\wedge I}\to T$$ such that $$H \circ in_0 = f, H\circ in_1 = g$$.
It's quite easy to prove : in one direction, if they're homotopic in the usual definition, then you may take $$S_{\wedge I} = S\times I$$ where $$I=[0,1]$$, $$in_i (s)= (s,i)$$ and $$p(s,t) = s$$ and then the usual homotopy $$H$$ works as an $$H$$.
Conversely, if $$f,g$$ are homotopic in this definition, then as $$p$$ is a homotopy equivalence, we get from $$p\circ in_0 = p\circ in_1$$ that $$in_0\sim in_1$$ so $$H\circ in_0\sim H\circ in_1$$ so $$f\sim g$$ : $$f$$ and $$g$$ are homotopic in the usual definition.