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We can regard $\Pi_{1}(X,x_{0})$ as the set of basepoint-preserving homotopy classes of maps $(S^{1},s_{0})\rightarrow(X,x_{0})$ . Let $[S^{1},X]$ be the set of homotopy classes of maps $S^{1}\rightarrow X$ , with no conditions on basepoints. Thus there is a natural map $\Phi:\Pi_{1}(X,x_{0})\rightarrow [S^{1},X]$ obtained by ignoring basepoints. Show that $\Phi$ is onto if $X$ is path-connected, and that $\Phi([f])= \Phi([g])$ iff $[f]$ and $[g]$ are conjugate in $\Pi_{1}(X,x_{0})$ . Hence $\Phi$ induces a one to one correspondence between $ [S^{1},X]$ and the set of conjugacy classes in $\Pi_{1}(X)$ when $X$ is path-connected.

this is problem 6 from hatcher page 38. for showing that it is onto,my Idea was to take $f$ in $[S^{1},X]$,and consider a point say $x\in f(S^{1})$,because of $X$ is path connected we can have a path from $x$ to $x_{0}$ call it $\alpha$,I know that it is not exact but I feel $f(S^{1})\cup \alpha$ is the element in $\Pi_{1}(X,x_{0})$ which we need.

for second part I have no Idea.

1.please help me to make them complete and exact.

2.I don't underestand this part well,when it say :"We can regard $\Pi_{1}(X,x_{0})$ as the set of basepoint-preserving homotopy classes of maps $(S^{1},s_{0})\rightarrow(X,x_{0})$" . please give me good imagination about it.

3.can you guide to how I must have a good vision to solve algebraic topology problems?I know by practicing it will be achieved but sometimes I feel that there is something that is more like a gap in my mind.

I will be very thankful for your consideration.

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