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Let $u:A\rightarrow B$ be a continuous map of topological spaces, $a\in A$, $b=u(a)$.

  1. How do I prove that there exists a unique group homomorphism $$u':\pi_1(A,a)\rightarrow\pi_1(B,b)$$ such that $u'([\gamma])=[u\circ\gamma]$ for all $\gamma\in P(A,a)$?

Let $v:B\rightarrow C$ be another continuous map, with $c=v(b)=(v\circ u)(a)$.

  1. How do I prove that for $${u':\pi_1(A,a)\rightarrow\pi_1(B,b)},\text{ }{v':\pi_1(B,b)\rightarrow\pi_1(C,c)},$$ $${(v\circ u)':\pi_1(A,a)\rightarrow\pi_1(C,c)}$$ we have $(v\circ u)'=v'\circ u'$?

What I know:

  1. $P(A,a)$ is the set of all closed paths with base point $a$. A closed path is a continuous map $\gamma:[0,1]\rightarrow A$ with $\gamma(0)=\gamma(1)=a$.
    The fundamental group $\pi_1(A,a)$ is the quotient of $P(A,a)$ to the equivalence relation $\sim$ of path-homotopy.
    How do I find a group homomorphism as above and how would I prove uniqueness?

  2. I'm certain that there is a 'trick' to this question which I just don't see. Also, I am fairly sure we can use the result of 1. for this question.

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  • $\begingroup$ Uniqueness is clear : every element has the form $[\gamma]$ and you specified what value it should take on these elements. What you have to check is that the value does not depend on the choice of $\gamma$, and that it's compatible with composition of paths. $\endgroup$ – Captain Lama Apr 27 '16 at 12:19
  • $\begingroup$ @CaptainLama How can I check this? And can you help with the second question? $\endgroup$ – user335050 Apr 27 '16 at 12:23
  • $\begingroup$ For the second question, $(v\circ u)'([\gamma]) = [(v\circ u)\circ \gamma] = [v \circ (u\circ \gamma)] = \dotsc$ $\endgroup$ – Daniel Fischer Apr 27 '16 at 12:39
  • $\begingroup$ @DanielFischer $...=[v(u'[\gamma])]=v'(u'[\gamma])=(v'\circ u')[\gamma]$? Is this correct? $\endgroup$ – user335050 Apr 27 '16 at 12:43
  • $\begingroup$ To whomever it may concern, why the downvote please? $\endgroup$ – user335050 Apr 27 '16 at 12:43
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So we have $u: A \rightarrow B$, continuous and $u(a) = b$ and we have $[\gamma] \in \pi_(A,a)$.

This means, as you say, that $\gamma: [0,1] \rightarrow A$ is continuous and $\gamma(0) = \gamma(1) = a$. Then $u \circ \gamma$ is continuous from $[0,1]$ into $C$, as a composition of continuous functions, and $$(u \circ \gamma)(0) = u(\gamma(0)) = u(a) = b = u(a) = u(\gamma(1)) = (u \circ \gamma)(1)$$

so it is a closed loop at $b$. Hence $[u \circ \gamma]$ is uniquely determined by the condition: there is no choice. So this defines $u'$, hopefully:

The only thing to check is well-definedness: $\gamma \sim \gamma'$, which is equivalent to $[\gamma'] = [\gamma]$, do we know that $u \circ \gamma \sim u \circ \gamma'$? If this is true (and only then) is the choice of which element of the class we compose with $u$ irrelevant for the value (as an equivalence class again).

The equivalence gives us a homotopy $H: [0,1] \times [0,1] \rightarrow A$ between $\gamma$ and $\gamma'$, which is a continuous function, such that $H(0,t) = H(1,t) = a$ for all $t$ (we keep endpoints fixed all the time) and $H(x,0) = \gamma(x)$ and $H(x,1) = \gamma'(x)$ (we move from $\gamma$ to $\gamma'$).

The homotopy to see the equivalence of $u \circ \gamma$ and $u \circ \gamma'$ is clearly $u \circ H$. One can easily check again (write it out!) that it fulfills the conditions, because $H$ does and $u$ maps $a$ to $b$. Continuity follows from composition of continuous functions.

Then you have to check that it is a group homomorphism, so take classes $[\gamma]$ and $[\gamma']$. Let $[\gamma] \ast [\gamma']$ be their product, see this write-up for all the details of checking the group structure. As a representative of this product we use the function

$$\gamma''(t) = \left\{ \begin{array}{ll} \gamma(2t) & \mbox{ if } 0 \le t \le \frac{1}{2}\\ \gamma'(2t-1) & \mbox{ if } \frac{1}{2} \le t \le 1 \end{array} \right.$$

Now $u'([\gamma]\ast[\gamma']) = u'([\gamma'']) = [u \circ g'']$. For this to be a homeomorphism, we need to see that this equals $u'([\gamma]) \ast u'([\gamma']) = [u \circ \gamma] \ast [u \circ \gamma']$. But writing down the representative of the latter in $\pi_1(B,b)$ in the same way, we get

$$(u \circ \gamma) \ast (u \circ \gamma')(t) = \left\{ \begin{array}{ll} u(\gamma(2t)) & \mbox{ if } 0 \le t \le \frac{1}{2}\\ u(\gamma'(2t-1)) & \mbox{ if } \frac{1}{2} \le t \le 1 \end{array} \right.$$

which is exactly $u \circ \gamma''$, and as the representative does not matter for the value (as a class), we have the required equality.

The second question is just writing out the definitions: for a class $[\gamma] \in \pi_1(A,a)$ we have:

$$(v \circ u)'([\gamma]) = [(v \circ u) \circ \gamma ] = [v \circ (u \circ \gamma)] = v'([u' \circ \gamma]) = v'(u'([\gamma])) = (v' \circ u')([\gamma])$$

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