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Algebraic Topology - Fundamental Group and the Homomorphism induced by h

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
On page 333 in Section 52: The Fundamental Group (Topology by Munkres) Munkres writes: (see attachement giving Munkres pages 333-334)

"Suppose that [tex] h: X \rightarrow Y [/tex] is a continuous map that carries the point [tex] x_0 [/tex] of X to the point [tex] y_0 [/tex] of Y.

We denote this fact by writing:

[tex] h: ( X, x_0) \rightarrow (Y, y_0) [/tex]

If f is a loop in X based at [tex] x_0 [/tex] , then the composite [tex] h \circ f : I \rightarrow Y [/tex] is a loop in Y based at [tex] y_0 [/tex]"

I am confused as to how this works ... can someone help with the formal mechanics of this.

To illustrate my confusion, consider the following ( see my diagram and text in atttachment "Diagram ..." )


Consider a point [tex] i^' [/tex] [tex] \in [0, 1][/tex] that is mapped by f into [tex] x^' [/tex] i.e. [tex] f( i^{'} ) [/tex] [tex] = x^' [/tex]

Then we would imagine that [tex] i^' [/tex] is mapped by [tex] h \circ f [/tex] into some corresponding point [tex] y^' [/tex] ( see my diagram and text in atttachment "Diagram ..." )


i.e. [tex] h \circ f (i^{'} ) [/tex] [tex] = y^' [/tex]

BUT

[tex] h \circ f (i^{'} ) = h(f(i^{'} )) = h(x^{'} ) [/tex]

But (see above) we only know of h that it maps [tex] x_0 [/tex] into [tex] y_0 [/tex]? {seems to me that is not all we need to know about h???}

Can anyone please clarify this situation - preferably formally and explicitly?

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
The definition of a loop in X (based at $x_0$) is that it is a continuous map $f:[0,1]\to X$ with $f(1) = f(0) = x_0$. So the point labelled $x_1$ in your diagram should actually be the same as $x_0$. The whole idea of a loop is that it goes round in a "circle" (though of course it needn't actually be circular) and comes back to where it started from.

If $f$ is a loop in $X$, based at $x_0$, and $h:(X,x_0)\to (Y,y_0)$ is continuous, then $h\circ f$ is a continuous map from [0,1] to $Y$ satisfying $(h\circ f)(1) = (h\circ f)(0) = y_0$, and that is all that is needed for $h\circ f$ so satisfy the definition of being a loop in $Y$ based at $y_0$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks Opalg.

Yes, can see I should have represented f as a loop

I have adjusted the Diagram ... see new attachement - Diagram 2 - f is a loop - ... ...

But then ... after such an adjustment my problem still remains and the previous argument still seems valid. To repeat the argument (still the same for a loop as a path I think ... BUT this time see Diagram 2 attached)

"Consider a point [tex] i^' [/tex] [tex] \in [0, 1][/tex] that is mapped by f into [tex] x^' [/tex] i.e. [tex] f( i^{'} ) [/tex] [tex] = x^' [/tex]

Then we would imagine that [tex] i^' [/tex] is mapped by [tex] h \circ f [/tex] into some corresponding point [tex] y^' [/tex] ( see my diagram and text in atttachment "Diagram ..." )


i.e. [tex] h \circ f (i^{'} ) [/tex] [tex] = y^' [/tex]

BUT

[tex] h \circ f (i^{'} ) = h(f(i^{'} )) = h(x^{'} ) [/tex]

But (see above) we only know of h that it maps [tex] x_0 [/tex] into [tex] y_0 [/tex]? {seems to me that is not all we need to know about h???} "


Obviously there is something I am missing ... can you clarify further ...

Peter
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I'm not sure that I understand the difficulty. The map $h:X\to Y$ is given. So each point $x'$ in $X$ gets mapped to a point $h(x')$ in $Y$. In this way, the whole loop in $X$ gets mapped to a loop in $Y$. The two loops may have very different shapes, but all that matters is that each of them is a continuous image of the unit interval whose endpoints coincide.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks again ... I suspect I am overthinking this issue ... I was looking for some formal rule assuring me that one actually achieved a loop in Y starting and finishing at [tex] y_0 [/tex] ... but I guess all you need to assure that you do get such a loop is continuity - continuity ensures the loop in X gets transformed into a loop (and not some other shape) in Y . Is that right?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I guess all you need to assure that you do get such a loop is continuity - continuity ensures the loop in X gets transformed into a loop (and not some other shape) in Y . Is that right?
Yes, all you need is continuity plus the endpoints going to the right place.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks again ... Appreciate your help

Peter