I am currently reading through Hatcher's Algebraic Topology book. I am having some trouble understanding the difference between a deformation retraction and just a retraction. Hatcher defines them as follows:

A deformation retraction of a space $X$ onto a subspace $A$ is a family of maps $f_t:X \to X$, $t \in I$, such that $f_0=\mathbb{1}$ (the identity map), $f_1(X)=A$, and $f_t|A=\mathbb{1}$ for all $t$.

A retraction of $X$ onto $A$ is a map $r:X \to X$ such that $r(X)=A$ and $r|A=\mathbb{1}$.

Is the notion of time the important characteristic that sets the two ideas apart? (It seems that the definition of deformation retraction utilizes time in its definition, whereas retraction seems to not.)

Any insight is appreciated. Also, if anyone have additional suggested reading material to help with concepts in Algebraic topology, that would be much appreciated.

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    $\begingroup$ the wikipedia page is very useful here (plus it mentions a possible terminological difference between hatcher and other authors) en.wikipedia.org/wiki/… $\endgroup$ – citedcorpse Sep 12 '13 at 21:58

The difference between a retraction and a deformation retraction does have to do with the "notion of time" as you suggest.

Here's a strong difference between the two:

1) For any $x_0 \in X$, $\{x_0\} \subset X$ has a retract. Choose $r : X \to \{x_0\}$ to be the unique map to the one-point set. Then, certainly, $r(x_0) = x_0$.

2) However, $\{x_0\} \subset X$ only has a deformation retraction if $X$ is contractible. To see, why, notice there has to be a family of maps $f_t : X \to X$ such that $f_0(x) = x$, $f_1(x) = x_0$, and $f_{t}(x_0) = x_0$ for every $t$. This gives a homotopy from $id_X$ to the constant map at $x_0$, which makes $X$ contractible.

In fact, showing a deformation retract from $X$ onto a subspace $A$ always exhibits that $A$ and $X$ are homotopy equivalent, whereas $A$ being a retract of $X$ is weaker. (But often, still useful! Two spaces being homotopy equivalent is very strong indeed!)

  • $\begingroup$ Thanks, @Thomas. Would it then be accurate to say that every deformation retraction is a retraction but not every retraction is a deformation retraction? $\endgroup$ – Tyler Clark Sep 12 '13 at 22:17
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    $\begingroup$ Yes; the map $f_1$ provided as part of the family of functions for a deformation retraction is enough to demonstrate that $A$ is a retract of $X$. $\endgroup$ – Thomas Belulovich Sep 12 '13 at 22:21
  • $\begingroup$ Oh, obviously! Thanks. I just wanted to make sure I was thinking about it appropriately. $\endgroup$ – Tyler Clark Sep 12 '13 at 22:28

As you noted the two notion are different, deformation retraction being a continuous family of continuous function, i.e. an homotopy, while a retraction being just a continuous function.

A retraction is just a map that sends all the point of $X$ in $A$ fixing the points of $A$.

A deformation retraction as the opposite is a family of mappings that fix the points of $A$, but that's more: we require that the family is continuous that means that we want that the induced map $X \times I \to X$ sending every pair $(x,t) \in X \times I$ to $f_t(x)$ is a continuous function.

Anyway there's also another way to see deformation retraction and more generally homotopies. For every space $Y$ we can consider the set $Y^I=\mathbf{Top}(I,Y)$ the set of continuous paths in $Y$ and topologizes this set with the compact-open topology.

Since the space $I$ is locally compact a general theorem tells that there's a bijection $$\mathbf{Top}(X \times I , Y) \cong \mathbf{Top}(X,Y^I)$$ sending every map $F \colon X \times I \to Y$ in the map $\bar F \colon X \to Y^I$ that to every $x \in X$ associates the continuous function $\bar F(x) \colon I \to Y$ such that for $t \in Y$ $\bar F(x)(t)=F(x,t)$ (this bijection in natural both in $X$ and $Y$).

Because of this bijection we can define an homotopy to be just a continuous function in $\mathbf{Top}(X,Y^I)$.

If we adopt this this point of view, of homotopies as continuous mapping in path spaces, a deformation retraction is just a mapping that associate to every point $x \in X$ a path, starting at the point $x$ and ending at some point of $A$. Each path corresponding to a point is the trajectory that the point follows during the deformation of $X$ to $A$.

Anyway a deformation retraction of $X$ to $A$ is not simply an homotopy: it is also an homotopy relative to the subspace $A$ between the identity and a map $r \colon X \to X$ such that $r(X) \subseteq A$. By the requirement that $r$ is homotopic to $1_X$ via an homotopy relative to $A$ it follows that $r$ must be a retraction: by definition an homotopy relative to $A$ sends every point of $A$ into the constant path which connect $a$ to $r(a)$, which must be equal.

This means that if $A$ is a deformation retract of $X$ then $A$ is also a retract, anyway the converse doesn't hold in general: a counterexample consider the map that send $S^1$ (the circle) to a point, this is a retraction, but there's no deformation retraction of $S^1$ to a point (to prove this one need to make a little work and build some invariant like the $\pi_1$ which is in the next chapter of the book).

Hope this helps understanding the ideas and the differences of these two concepts.


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