# Algebraic Topology - Fundamental Group and the Homomorphism induced by h

#### Peter

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

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

We denote this fact by writing:

$$h: ( X, x_0) \rightarrow (Y, y_0)$$

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

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

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

i.e. $$h \circ f (i^{'} )$$ $$= y^'$$

BUT

$$h \circ f (i^{'} ) = h(f(i^{'} )) = h(x^{'} )$$

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

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

Peter

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#### Opalg

##### MHB Oldtimer
Staff member
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

#### Peter

##### Well-known member
MHB Site Helper
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 $$i^'$$ $$\in [0, 1]$$ that is mapped by f into $$x^'$$ i.e. $$f( i^{'} )$$ $$= x^'$$

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

i.e. $$h \circ f (i^{'} )$$ $$= y^'$$

BUT

$$h \circ f (i^{'} ) = h(f(i^{'} )) = h(x^{'} )$$

But (see above) we only know of h that it maps $$x_0$$ into $$y_0$$? {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

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#### Opalg

##### MHB Oldtimer
Staff member
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

#### Peter

##### Well-known member
MHB Site Helper
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 $$y_0$$ ... 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
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

#### Peter

##### Well-known member
MHB Site Helper
Thanks again ... Appreciate your help

Peter