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I'm pretty sure this has an easy solution, but I can't seem to find it.

Let $X$ be a contractible $2$-dimensional CW-complex, let $\gamma$ be an embedded loop in $X$, and let $f : D^2 \rightarrow X$ be an embedding of a disc in $X$ which maps the boundary of $D$ to $\gamma$.

My question is the following. Let $f' : D^2 \rightarrow X$ be a continuous map of a disc into $X$ which takes the boundary of $D$ to $\gamma$. Must we then have $f(D^2) \subset f'(D^2)$ ? I'm pretty sure that the answer is yes, but I can't seem to prove it.

Of course, this has an obvious generalization to higher dimensional complexes, and I'd be interested in that too.

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I guess at least you mean $f'$ is also injective –  Yi Liu Aug 19 '10 at 4:15
    
@Yi Liu: There is no need to assume $f'$ is injective. –  Tom Church Aug 19 '10 at 4:28
    
@Tom: Oh you're right. I thought the inclusion is the other way around. never mind! –  Yi Liu Aug 19 '10 at 4:46
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up vote 3 down vote accepted

One way I can think of is to take a point $x\in f(D)\setminus f'(D)$, assuming on the contrary. One may assume $x$ lies in the interior of $f(D)$ and the interior of some 2-cell. Then you can remove a small disk $U$ in $f(D)$ which still lies in the 2-cell, and a M-V sequence argument shows $\gamma$ is homologically nontrivial in $H_1(X\setminus U)$, as $[\gamma]=[\partial U]$. This gives a contradiction.

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